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9-5 The Distance Formula
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Electrostatics The section of CBSE Class 12 Physics electrostatic potential and capacitance notes mainly deals with the in-depth analysis of electromagnetic phenomena when they are not performing any movements. Additionally, it is divided into ten further sub-topics to study the companion processes of reaching the state. These are - 1. Electric charge In this section of Physics ch 2 Class 12 notes, you get to learn about the basic features of electric charge and its expression in Physics. Along with its basics, the sections help to understand the full potential of charge. Different aspects of Charge included in Class 12 Physics Chapter 2 notes are - Definition Type: Positive and Negative Charge Unit and dimensional formula Point Charge Properties of Charge Comparison of Charge and Mass Methods of Charging Electroscope 2. Coulomb's Law Force is created when charges of opposite signs attract each other, and they repulse if the signs are the same. Coulomb's law tries to define this phenomenon through a mathematical formula, explicitly mentioned in Physics Class 12 notes Chapter 2. Moreover, there is key information about the variation of the constant k and its effect on a medium. Coulomb's law's vector form and the principle of superimposition are also explained in ch 2 Physics Class 12 notes. (Image will be uploaded soon) 3. Electric Field As stated in Class 12 Physics Chapter 2 notes, every positively or negatively charged particle has their respective electric fields. It feels a force at the time of interaction which might be attraction or repulsion. As it arises from electric charge, it is crucial to know about its different parts like - Electric field intensity Relation between electric force and electric field Super imposition of electric field Point charge Continuous charge distributions Properties of Electric Field Lines Motion of Charged Particles in an Electric field Learning more about the electric field from electric potential and capacitance notes Class 12 helps a student to get a grasp of upcoming chapters. 4. Electric Potential Energy When energy helps a charge to move from an electric field, it is known as the Electric Potential Energy. This section of electrostatic chapter Class 12 notes requires a student to study the Electron volt (eV), and the potential energy that an n number of charges can hold. 5. Electric Potential This section of Class 12 Physics Chapter 2 notes focuses on in-depth learning of Electric Potential or Voltage. Basically, it defines the potential movement of energy. 6. Relation between Electric Field and Potential Apart from knowing more about the relationship between the two values, Physics Class 12 Chapter 2 notes also discuss equipotential surfaces. 7. Electric Dipole Essentially, 'Dipoles' are two opposite points of charge represented with q and –q, with their distance between each other being 2a. Electric Dipoles are crucial in your study of Physics Class 12 Chapter 2 notes to learn more about electric fields and their potential. Additionally, Class 12 Physics Chapter 2 notes focus on the influence of electric dipoles on a uniform electric field mainly through Force and Torque, Work, and Potential Energy. In the last part of Electrostatics, further focus is on using the formulas to their fullest potential. It includes subsections of Electric Field, Electric Potential Energy, Electric Potential, and Electric Dipole. In the notes for electrostatic potential and capacitance, you will find proper solutions accompanied by clear and crisp diagrams for better understanding. 8. Gauss's Law Apart from just discussing the Gauss's Law, in Physics Class 12 ch 2 notes there is a thorough explanation of its properties and applications. The Gauss' Law states that net electric flux passing through a hypothetical closed surface is equal to the net electric charge present within the same closed surface. Being a broad part of the whole chapter, you may need to spend a little more time on it. Moving forward, it starts discussing the properties of conductors in relation to Gauss's Law. The Class 12 Physics notes Chapter 2 perfectly defines the journey to Gauss' Law from Coulomb's Law. Here is the Gauss's Law present in the Class 12 Physics ch 2 notes, (image will be uploaded soon) 9. Capacitors There is a dedicated section about Capacitors in the Class 12 Physics Chapter 2 notes elucidating its functions and importance as storage of potential electric energy. After explaining the structure of a capacitor, it points out the different types, parallel plate, spherical and cylindrical. The section of Chapter 2 notes of Physics Class 12 is further divided into subheads like: Properties of an ideal battery Grouping of capacitors Simple circuits (Series and Parallel) Dielectric Van de Graaff generator Combination of drops Charge distribution method Wheatstone Bridge-based circuit Extended Wheatstone Bridge Infinite network of capacitors Redistribution of charge between two capacitors Vedantu prepares the Class 12 Physics Chapter 2 notes with help from subject matter experts. In the PDF, you get a comprehensive idea of the topic along with potential answers to the most asked questions. Furthermore, the detailed explanation on each section and subsections are written in a simple language allows a student to ace their exams with wholesome knowledge. These Physics Chapter 2 Class 12 notes are going to be one of the best supplementary study materials besides a student’s textbooks. Visit the Vedantu website or download the app to get your hands on all important notes! Important Questions A charge of 4 × 10–8C is uniformly distributed on the surface of a spherical conductor, having a radius of 15 cm. Determine the electric field just outside this sphere at a point that is 15 cm from the centre of this sphere. Determine the capacitance given that the distance between the two plates has been reduced by half and the parallel plate capacitor holds a capacitance of 20 pF (where 1pF = 10-12 F) having air between the two plates. What will be the total capacitance of a combination where three capacitors, each having a capacitance of 20 pF, are connected in series. A square having a side of 10 cm has a 500 µC charge at its centre. Determine the work done to move a charge of 10 µC between two points that are diagonally opposite each other on the square. At an equatorial point, what will be the electrostatic potential because of an electric dipole? Calculate the work done to move a test charge, q, through a length of 1 cm along the equatorial axis of an electric dipole? Polarisation A capacitor has its plates enclosed in a medium that can be filled by insulating substances. A net dipole moment is then induced by an electric field in the dielectric. This event causes the field in an opposite direction. Equipotential Surface An equipotential surface is a type of surface where the potential always has a constant value. If considered as a point charge, the concentric spheres that are centred at a particular area of this charge are basically equipotential surfaces. Advantages of Vedantu's Revision Notes: A Comprehensive Resource for Effective Learning There are several reasons why one may refer to Vedantu's revision notes for studying a subject like Electrostatic Potential and Capacitance. Here are some key points: Comprehensive Coverage: Vedantu's revision notes provide a comprehensive coverage of the entire topic, ensuring that all important concepts and subtopics are included. Concise and Organized: The notes are designed to be concise, focusing on the key points and core ideas. They are organized in a structured manner, making it easy for students to navigate and revise the content. Simplified Explanation: The revision notes offer simplified explanations of complex concepts, making them more accessible and easier to understand. This helps students grasp the material more effectively. Key Formulas and Equations: The notes highlight the key formulas and equations relevant to the topic, ensuring that students have a clear understanding of the mathematical aspects of Electrostatic Potential and Capacitance. Examples and Illustrations: Vedantu's revision notes often include examples and illustrations that help clarify concepts and provide practical applications, enabling students to better relate theory to real-world scenarios. Quick Recap: The revision notes serve as a quick recap of the important points, allowing students to review the material efficiently before exams or assessments. Exam-Oriented Approach: Vedantu's revision notes are designed with an exam-oriented approach, focusing on the topics and concepts that are frequently asked in examinations. This helps students prepare effectively and increase their chances of scoring well. Accessible Anytime: Vedantu's revision notes are easily accessible online, allowing students to study at their convenience and revise the material anytime, anywhere.
Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an n value, where n = 1, 2, 3, …. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 9.4.1). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: The values nf and ni are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital. Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction ψ is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is n – l – 1. Consider the examples in Figure 9.4.2. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 – 0 – 1 = 1 node for 2s, and 3 – 0 – 1 = 2 nodes for the 3s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY If an electron has an angular momentum (l ≠ 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to –1, 0, or +1. Generally speaking, ml can be equal to –l, –(l – 1), …, –1, 0, +1, …, (l – 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure 9.4.3. Figure 9.4.4 illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ≥ 1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms. The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron “rotation” or “spinning.” Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number ms=12. The other is called the β state, with the z component of the spin being negative and ms=−12. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having ms=−12 and ms=12 are different if an external magnetic field is applied. Figure 9.4.5 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the 12 spin quantum number and down (in the negative z direction) for the spin quantum number of −12. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with ms=12 has a slightly lower energy in an external field in the positive z direction, and an electron with ms=−12 has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values (±12), no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly
1. Absolute Value - The magnitude of a number or the distance from 0 on a real number line. 2. Addend - Any number or quantity being added. (addend + addend = sum) 3. Algorithm - Step-by-step process to accomplish a task. 4. Difference - The amount that remains after one quantity is subtracted from another. (The answer to a subtraction problem) 5. Dividend - A quantity or total to be divided. (dividend ÷ divisor = quotient) 6. Divisor - The number per group by which the dividend is to be divided. 7. Factor - A number that is multiplied by another number. 8. Integers - A set of numbers that includes natural numbers, their opposites, and zero. (… -3, -2, -1, 0, 1, 2, 3 …) 9. Product - The result of a multiplication expression (the answer to a multiplication problem) 10. Quotient - The result of the division of one quantity by another (the answer to a division problem) 11. Rational Numbers - Numbers that can be expressed
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.
1. [Force] Part A: A student wants to test how friction affects a toy car. She rolls the car across a sheet of sandpaper and then across a sheet of wax paper. Which is the independent (changing) variable? A. The speed of the car B. The type of surface C. The distance traveled D. The size of the car Part B: On which surface will the car likely stop the SOONEST? A. The wax paper B. The sandpaper C. Both will be the same D. Neither surface has friction 2. [Magnets] Which of these is a measurable question for a magnet experiment? A. Are magnets more fun than springs? B. What is the prettiest color for a magnet? C. How many steel paperclips can a bar magnet lift? D. Why were magnets invented? 3. [Earth's Changes] A student observes a statue in a park that has lost its nose and has smooth edges after many years of rain and wind. What process caused this? A. Erosion B. Deposition C. Weathering D. Evaporation 4. [Earth's Changes] When a river reaches the ocean, it slows down and creates a landform called a delta by dropping sand and silt. This "dropping off" is called: A. Weathering B. Deposition C. Condensation D. Friction 5. [Resources] Why is coal considered a nonrenewable resource? A. It can be burned to make electricity. B. It is found deep underground. C. It takes millions of years to form and cannot be replaced quickly. D. It is made from ancient plants. 6. [Conservation] A school replaces all its old lightbulbs with energy-efficient LED bulbs. This is an example of: A. Weathering a resource B. Conserving a resource C. Deposition of energy D. Creating a renewable resource 7. [Aquifers] An aquifer is like a giant underground sponge. What characteristic of the rocks allows them to hold water? A. The rocks are solid and water-proof. B. The rocks are porous, with tiny spaces for water to sit. C. The rocks are magnetic and pull water toward them. D. The rocks are melted into a liquid state. 8. [Water Cycle] On a humid morning, you see dew on the grass even though it didn't rain overnight. Which part of the water cycle formed the dew? A. Evaporation B. Precipitation C. Condensation D. Transpiration 9. [Climate] Which of the following is a description of CLIMATE? A. "It is currently 85 degrees in McAllen." B. "There is a 40% chance of rain this afternoon." C. "South Texas typically has mild winters and very hot summers." D. "The wind is blowing from the North at 10 mph today." 10. [Weather/Climate] A scientist is looking at a chart that shows the total annual rainfall in a city from 1990 to 2020. What is the scientist most likely studying? A. The daily weather forecast B. The climate of the region C. The water cycle of a single pond D. The rate of erosion on a local hill
Owls, such as the young snowy owls on the previous page, have for centuries been symbols of both wisdom and mystery. To many cultures their piercing eyes have conveyed a look of intelligence. Their silent flight through darkened landscapes in search of prey has projected an air of power or wonder. For this chapter and this book, owls are an engaging example of a living organism from the world of biology—the study of life. BIOLOGY AND YOU Living in a small town, in the country, or at the edge of the suburbs, one may be lucky enough to hear an owl's hooting. This experience can lead to questions about where the bird lives, what it hunts, and how it finds its prey on dark, moonless nights. Biology, or the study of life, offers an organized and scientific framework for posing and answering such questions about the natural world. Biologists study questions about how living things work, how they interact with the environment, and how they change over time. Biologists study many different kinds of living things ranging from tiny organisms, such as bacteria, to very large organisms, such as elephants. Each day, biologists investigate subjects that affect you and the way you live. For example, biologists determine which foods are healthy. As shown in Figure 1-1, everyone is affected by this impor- tant topic. Biologists also study how much a person should exer- cise and how one can avoid getting sick. Biologists also study what CHARACTERISTICS OF LIFE The world is filled with familiar objects, such as tables, rocks, plants, pets, and automobiles. Which of these objects are living or were once living? What are the criteria for assigning something to the living world or the nonliving world? Biologists have established that living things share seven characteristics of life. These characteristics are organization and the presence of one or more cells, response to a stimulus (plural, stimuli), homeostasis, metabolism, growth and development, reproduction, and change through time. Organization and Cells Organization is the high degree of order within an organism’s internal and external parts and in its interactions with the living world. For example, compare an owl to a rock. The rock has a spe- cific shape, but that shape is usually irregular. Furthermore, differ- ent rocks, even rocks of the same type, are likely to have different shapes and sizes. In contrast, the owl is an amazingly organized individual, as shown in Figure 1-2. Owls of the same species have the same body parts arranged in nearly the same way and interact with the environment in the same way. Copyright © by Holt, Rinehart and Winston. All rights reserved. ORGANISM (Barn Owl) ORGAN (Owl’s Ear) TISSUE (Nervous Tissue Within the Ear) CELL (Nerve Cell) your air, land, and fAll living organisms, whether made up of one cell or many cells, have some degree of organization. A cell is the smallest unit that can perform all life’s processes. Some organisms, such as bacteria, are made up of one cell and are called unicellular (YOON-uh-SEL-yoo-luhr) organisms. Other organisms, such as humans or trees, are made up of multiple cells and are called multicellular (MUHL-ti-SEL-yoo-luhr) organisms. Complex multicellular organisms have the level of orga- nization shown in Figure 1-2. In the highest level, the organism is made up of organ systems, or groups of specialized parts that carry out a certain function in the organism. For example, an owl’s ner- vous system is made up of a brain, sense organs, nerve cells, and other parts that sense and respond to the owl’s surroundings. Organ systems are made up of organs. Organs are structures that carry out specialized jobs within an organ system. An owl’s ear is an organ that allows the owl to hear. All organs are made up of tissues. Tissues are groups of cells that have similar abilities and that allow the organ to function. For example, nervous tissue in the ear allows the ear to detect sound. Tissues are made up of cells. A cell must be covered by a membrane, contain all genetic information necessary for replication, and be able to carry out all cell functions. Within each cell are organelles. Organelles are tiny structures that carry out functions necessary for the cell to stay alive. Organelles contain biological molecules, the chemical compounds that provide physical structure and that bring about movement, energy use, and other cellular functions. All biological molecules are made up of atoms. Atoms are the simplest particle of an ele- ment that retains all the properties of a certain element. Response to Stimuli Another characteristic of life is that an organism can respond to a stimulus—a physical or chemical change in the internal or external environment. For example, an owl dilates its pupils to keep the level of light entering the eye constant. Organisms must be able to respond and react to changes in their environment to stay alive. ORGANELLE (Mitochondrion) BIOLOGICAL MOLECULE (Phospholipid) ATOM (Oxygen) cell from the Latin, cella meaning “small room,” or “hut” Word Roots and Origins www.scilinks.org Topic: Characteristics of Life Keyword: HM60257 mb06se_bios01.qxd 5/18/07 10:37 AM Page 7 8 CHAPTER 1 Homeostasis All living things, from single cells to entire organisms, have mecha- nisms that allow them to maintain stable internal conditions. Without these mechanisms, organisms can die. For example, a cell’s water content is closely controlled by the taking in or releas- ing of water. A cell that takes in too much water will rupture and die. A cell that doesn’t get enough water will also shrivel and die. Homeostasis (HOH-mee-OH-STAY-sis) is the maintenance of a stable level of internal conditions even though environmental conditions are constantly changing. Organisms have regulatory systems that maintain internal conditions, such as temperature, water content, and uptake of nutrients by the cell. In fact, multi- cellular organisms usually have more than one way of maintain- ing important aspects of their internal environment. For example, an owl’s temperature is maintained at about 40°C (104°F). To keep a constant temperature, an owl’s cells burn fuel to produce body heat. In addition, an owl’s feathers can fluff up in cold weather. In this way, they trap an insulating layer of air next to the bird’s body to maintain its body temperature. Metabolism Living organisms use energy to power all the life processes, such as repair, movement, and growth. This energy use depends on metabolism (muh-TAB-uh-LIZ-uhm). Metabolism is the sum of all the chemical reactions that take in and transform energy and materials from the environment. For example, plants, algae, and some bacteria use the sun’s energy to generate sugar molecules during a process called photosynthesis. Some organisms depend on obtaining food energy from other organisms. For instance, an owl’s metabolism allows the owl to extract and modify the chemi- cals trapped in its nightly prey and use them as energy to fuel activities and growth. Growth and Development All living things grow and increase in size. Some nonliving things, such as crystals or icicles, grow by accumulating more of the same material of which they are made. In contrast, the growth of living things results from the division and enlargement of cells. Cell division is the formation of two new cells from an existing cell, as shown in Figure 1-3. In unicellular organisms, the primary change that occurs following cell division is cell enlargement. In multi- cellular life, however, organisms mature through cell division, cell enlargement, and development. Development is the process by which an organism becomes a mature adult. Development involves cell division and cell differen- tiation, or specialization. As a result of development, an adult organism is composed of many cells specialized for different func- tions, such as carrying oxygen in the blood or hearing. In fact, the human body is composed of trillions of specialized cells, all of which originated from a single cell, the fertilized egg. This unicellular organism, Escherichia coli, inhabits the human intestines. E. coli reproduces by means of cell division, during which the original cell splits into two identical offspring cells. FIGURE 1-3 Observing Homeostasis Materials 500 mL beakers (3), wax pen, tap water, thermometer, ice, hot water, goldfish, small dip net, watch or clock with a second hand Procedure 1. Use a wax pen to label three 500 mL beakers as follows: 27°C (80°F), 20°C (68°F), 10°C (50°F). Put 250 mL of tap water in each beaker. Use hot water or ice to adjust the tem- perature of the water in each beaker to match the temperature on the label. 2. Put the goldfish in the beaker of 27°C water. Record the number of times the gills move in 1 minute. 3. Move the goldfish to the beaker of 20°C water. Repeat observations. Move the goldfish to the beaker of 10°C. Repeat observations. Analysis What happens to the rate at which gills move when the temp- erature changes? Why? How do gills help fish maintain homeostasis? Quick Lab mb06se_bios01.qxd 5/18/07 10:37 AM Page 8 THE SCIENCE OF LIFE 9 Reproduction All organisms produce new organisms like themselves in a process called reproduction. Reproduction, unlike other characteristics, is not essential to the survival of an individual organism. However, because no organism lives forever, reproduction is essential for the continuation of a species. Glass frogs, as shown in Figure 1-4, lay many eggs in their lifetime. However, only a few of the frogs’ off- spring reach adulthood and successfully reproduce. During reproduction, organisms transmit hereditary informa- tion to their offspring. Hereditary information is encoded in a large molecule called deoxyribonucleic acid, or DNA. A short segment of DNA that contains the instructions for a single trait of an organism is called a gene. DNA is like a large library. It contains all the books—genes—that the cell will ever need for making all the struc- tures and chemicals necessary for life. Hereditary information is transferred to offspring during two kinds of reproduction. In sexual reproduction, hereditary information recombines from two organisms of the same species. The resulting offspring are similar but not identical to their parents. For example, a male frog’s sperm can fertilize a female’s egg and form a single fer- tilized egg cell. The fertilized egg then develops into a new frog. In asexual reproduction, hereditary information from different organisms is not combined; thus the original organism and the new organism are genetically the same. A bacterium, for example, reproduces asexually when it splits into two identical cells. Change Through Time Although individual organisms experience many changes during their lifetime, their basic genetic characteristics do not change. However, populations of living organisms evolve or change through time. The ability of populations of organisms to change over time is important for survival in a changing world. This factor is also impor- tant in explaining the diversity of life-forms we see on Earth today. 1. How does biology affect a person’s daily life? 2. How does biology affect society? 3. Name the characteristics shared by living things. 4. Summarize the hierarchy of organization found in complex multicellular organisms. 5. What are the different functions of homeostasis and metabolism in living organisms? 6. How does the growth among living and nonliv- ing things differ? 7. Why is reproduction an important characteristic of life? CRITICAL THINKING 8. Applying Information Crystals of salt grow and are highly organized. Why don’t biologists con- sider them to be alive? 9. Analyzing Models When a scientist designs a space probe to detect life on a distant planet, what kinds of things should it measure? 10. Making Comparisons Both cells and organisms share the characteristics of life. How are cells and organismsood supply will be like in the near future.EVOLUTION OF LIFE Individual organisms change during their lifetime, but their basic genetic characteristics do not change. However, populations of liv- ing organisms do change through time, or evolve. Evolution, or descent with modification, is the process in which the inherited characteristics within populations change over generations, such that genetically distinct populations and new species can develop. Evolution as a theme in biology helps us understand how the various branches of the “tree of life” came into existence and have changed over time. It also explains how organisms alive today are related to those that lived in the past. Finally, it helps us understand the mechanisms that underlie the way organisms look and behave. Natural Selection The ability of populations of organisms to change over time is important for survival in a changing world. According to the theory of evolution by natural selection, organisms that have certain favorable traits are better able to survive and reproduce success- fully than organisms that lack these traits. One product of natural selection is the adaptation of organisms to their environment. Adaptations are traits that improve an indi- vidual’s ability to survive and reproduce. For example, rabbits with white fur and short ears in a snowy place, such as the one in Figure 1-7a, may avoid predators and frostbitten ears more often than those with dark fur and long ears. Thus, the next generation of rabbits will have a greater percentage of animals carrying the genes for white fur and short ears. In contrast, the brown, long- eared rabbit, as shown in Figure 1-7b, would survive and reproduce more successfully in a hot desert environment. The survival and reproductive success of organisms with favor- able traits cause a change in populations of organisms over gener- ations. This descent with modification is an important factor in explaining the diversity of organisms we see on Earth today. 1. Name three unifying themes found in biology. 2. How is the unity and diversity in the living world represented? 3. Identify the three domains and the kingdoms found in each domain. 4. How are organisms interdependent? 5. Describe why evolution is important in explain- ing the diversity of life. 6. Distinguish between evolution and natural selection. CRITICAL THINKING 7. Applying Information Assign the various top- pings you put on pizza to the appropriate domains and kingdoms of life. 8. Analyzing Graphics According to the “tree” in Figure 1-5, which of these pairs are more closely related: Archaea:Bacteria or Archaea:Eukarya? 9. Making Hypotheses Fossil evidence shows that bats descended from shrewlike organisms that could not fly. Write a hypothesis for how natural selection might have led to flying bats. SECTION 2 REVIEW (a) This short-eared arctic hare, Lepus arcticus, is hidden from predators and protected from frostbite in a snowy environment. (b) The mottled brown coats of desert rabbits blend in with the dirt and dry grasses, and their long ears help them radiate excess heat and thus avoid overheating. FIGURE 1-7 (a) (b) Copyright © by Holt, Rinehart and Winston. All rights reserved. THE SCIENCE OF LIFE 13 TH E STUDY OF BIOLOGY Curiosity leads us to ask questions about life. Science provides a way of answering such questions about the natural world. Science is a systematic method that involves forming and testing hypotheses. More importantly, science relies on evidence, not beliefs, for drawing conclusions. SCIENCE AS A PROCESS Science is characterized by an organized approach, called the scientific method, to learn how the natural world works. The methods of science are based on two important principles. The first principle is that events in the natural world have natural causes. For example, the ancient Greeks believed that lightning and thunder occurred because a supernatural god Zeus hurled thunderbolts from the heavens. By contrast, a scientist considers lightning and thunder to result from electric charges in the atmos- phere. When trying to solve a puzzle from nature, all scientists, such as the one in Figure 1-8, accept that there is a natural cause to solve that puzzle. A second principle of science is uniformity. Uniformity is the idea that the fundamental laws of nature operate the same way at all places and at all times. For example, scientists assume that the law of gravity works the same way on Mars as it does on Earth. Steps of the Scientific Method Although there is no single method for doing science, scientific studies involve a series of common steps. 1. The process of science begins with an observation. An observation is the act of perceiving a natural occurrence that causes someone to pose a question. 2. One tries to answer the question by forming hypotheses (singular, hypothesis). A hypothesis is a proposed explanation for the way a particular aspect of the natural world functions. 3. A prediction is a statement that forecasts what would happen in a test situation if the hypothesis were true. A prediction is recorded for each hypothesis. 4. An experiment is used to test a hypothesis and its predictions. 5. Once the experiment has been concluded, the data are analyzed and used to draw conclusions. 6. After the data have been analyzed, the data and conclusions are communicated to scientific peers and to the public. This way oth- ers can verify, reject, or modify the researcher’s conclusions. SECTION 3 OBJECTIVES ● Outline the main steps in the scientific method. ● Summarize how observations are used to form hypotheses. ● List the elements of a controlled experiment. ● Describe how scientists use data to draw conclusions. ● Compare a scientific hypothesis and a scientific theory. ● State how communication in science helps prevent dishonesty and bias. VOCABULARY scientific method observation hypothesis prediction experiment control group experimental group independent variable dependent variable theory peer review All researchers, such as the one releasing an owl above, use the scientific method to answer the questions they have about nature. FIGURE 1-8 Copyright © by Holt, Rinehart and Winston. All rights reserved. 14 CHAPTER 1 OBSERVING AND ASKING QUESTIONS The scientific method generally begins with an unexplained observa- tion about nature. For example, people have noticed for thousands of years that owls can catch prey in near total darkness. As shown in steps and of Figure 1-9, an observation may then raise ques- tions. The owl observation raises the question: How does an owl detect prey in the dark? FORMING A HYPOTHESIS After stating a question, a biologist lists possible answers to a sci- entific question—hypotheses. Good hypotheses answer a question and are testable in the natural world. For example, as shown in step Figure 1-9, there are several possible hypotheses for the question of how owls hunt at night: (a) owls hunt by keen vision in the dark; (b) owls hunt by superb hearing; or (c) owls hunt by detecting the prey’s body heat. Predicting To test a hypothesis, scientists make a prediction that logically fol- lows from the hypothesis. A prediction is what is expected to hap- pen if each hypothesis were true. For example, if hypothesis (a) is true, (owls hunt by keen night vision) then one can predict that the owl will pounce only on the mouse in either a light or a dark room. If hypothesis (b) is true (owls hunt by hearing), then one can pre- dict that in a lighted room, the owl will pounce closer to the mouse’s head. But, in a dark room, the owl should pounce closer to a rustling leaf attached to the mouse. Finally, if hypothesis (c) is true (owls hunt by sensing body heat), then an owl would strike only the prey no matter the room conditions, because owls hunt by detecting the prey’s body heat. 3 1 2 Copyright © by Holt, Rinehart and Winston. All rights reserved. A scientific study includes observations, questions, hypotheses, predictions, experiments, data analysis, and conclu- sions. A biologist can use the scientific method to set up an experiment to learn how an owl captures prey at night. FIGURE 1-9 1 OBSERVATION Owls capture prey on dark nights. 2 QUESTION How do owls detect prey on dark nights? 3 HYPOTHESES a) Owls hunt in the dark by vision. b) Owls hunt in the dark by hearing. c) Owls hunt in the dark by sensing body heat. THE SCIENCE OF LIFE 15 Notice that these predictions make it difficult to distinguish be- tween the vision and body heat hypotheses. The reason is that both hypotheses predict that the owl could grab the mouse in a dark room. Also, these three hypotheses do not eliminate all other factors that could influence how the owl finds its prey. However, testing predictions can allow one to begin rejecting hypotheses and thus to get closer to determining the answer(s) to a question. DESIGNING AN EXPERIMENT Biologists often test hypotheses by setting up an experiment. Step in Figure 1-9 outlines an experiment to test the hypotheses about how an owl hunts at night. First, experimenters set up a room with an owl perch high on one side and a small trap door on the other side for releasing mice. Then, they tied a leaf to each mouse’s tail with a string and released each mouse into the room. Next, each mouse ran silently across the room, but the leaf trailed behind, making a rustling noise. During half of the trials, the lights were on. During the other half, the room was dark. Technicians videotaped all the action in the chamber with an infrared light, which owls cannot see. The researchers then viewed the videos and measured the position of the owl’s strike relative to each mouse’s head. Performing the Experiment Many scientists use a controlled experiment to test their hypotheses. A controlled experiment compares an experimental group and a control group and only has one variable. The control group pro- vides a normal standard against which the biologist can compare results of the experimental group. The experimental group is iden- tical to the control group except for one factor, the independent variable. The experimenter manipulates the independent variable, sometimes called the manipulated variable. 4 4 EXPERIMENT 5 DATA COLLECTION AND ANALYSIS Measure and compare the distance from the owl’s strike to the mouse and to the leaf in light and dark. 6 CONCLUSION Data supported the hearing hypothesis: Owls hunt in the dark by hearing. prey Test predictions of the three hypotheses. Control: In the light Experimental: In the dark 1 2 3 4 5 6 7 8 9 10 11 Predicting Results Materials 2 Petri dishes with agar, cellophane tape, wax pen Procedure 1. Open one of the Petri dishes, and streak your finger across the surface of the agar. 2. Replace the lid, and seal it with the tape. Label this Petri dish with your name and a number 1. 3. Seal the second Petri dish with- out removing the lid. Label this Petri dish with your name and the number 2. 4. Write a prediction about what will happen in each dish. Store your dishes as your teacher directs. Record your observations. Follow your teacher’s directions for disposal of your dishes. Analysis Was your prediction accurate? What evidence can you cite to support your prediction? If you did not obtain the results you predicted, would you change your testing method or your prediction? Explain. Evaluate the importance of obtaining a result that does not support your prediction. Quick Lab mb06se_bios03.qxd 5/18/07 10:40 AM Page 15 16 CHAPTER 1 The independent variable in the owl experiment is the presence or absence of light. In the owl experiment, the control group hunts in the light, and the experimental group hunts in the dark. In addi- tion to varying the independent variable, a scientist observes or measures another factor called the dependent variable, or respond- ing variable, because it is affected by the independent variable. In the owl experiment, the dependent variable is distance from the owl’s strike to the mouse’s head. Testing the Experiment Some controlled experiments are conducted “blind.” In other words, the biologist who scores the results is unaware of whether a given subject is part of the experimental or control group. This factor helps eliminate experimenter bias. Experiments should also be repeated, because living systems are variable. Moreover, scien- tists must collect enough data to find meaningful results. COLLECTING AND ANALYZING DATA Most experiments measure a variable—the dependent variable. This measurement provides quantitative data, data measured in numbers. For example, in the experiment above, scientists mea- sured the distance of an owl’s strike from the prey’s head in cen- timeters, as shown in step of Figure 1-9. An event’s duration in milliseconds is also an example of quantitative data. Biologists usually score the results of an experiment by using one of their senses. They might see or hear the results of an experiment. Scientists also extend their senses with a micro- scope for tiny objects or a microphone for soft sounds. In the owl experiment, biologists extended their vision with infrared cameras. Analyzing and Comparing Data After collecting data from a field study or an experiment and then organizing it, biologists then analyze the data. In analyzing data, the goal is to determine whether the data are reliable, and whether they support or fail to support the predictions of the hypothesis. To do so, scientists may use statistics to help determine relation- ships between the variables involved. They can then compare their data with other data that were obtained in other similar studies. It is also important at this time to determine possible sources of error in the experiment just per- formed. Scientists usually display their data in tables or graphs when analyzing it. For the owl study, biologists could have made a bar graph such as the one in Figure 1-10, which shows the average distance from the owl’s strike relative to the mouse’s head or the leaf in the light and in the dark. 5 5 0 10 15 20 25 In the light In the dark Average distance from strike (cm) Distance Between Owl Strike and a Mouse or From a Leaf Attached to Mouse 30 Mouse Leaf Mouse Leaf The data below are hypothetical results that might occur from the described owl experiment.The independent variable is the darkness of the room, and the dependent variable is how far the owl struck from the mouse’s head.The data show that the owl strikes more accurately at the mouse in the light but strikes more accurately at the leaf in the dark. FIGURE 1-10 Copyright © by Holt, Rinehart and Winston. All rights reserved. THE SCIENCE OF LIFE 17 DRAWING CONCLUSIONS Biologists analyze their tables, graphs, and charts to draw conclu- sions about whether or not a hypothesis is supported, as shown in step of Figure 1-9. The hypothetical owl data show that in the light, owls struck with greater accuracy at the mouse than at the leaf, but in the dark, owls struck with greater accuracy at the leaf than the mouse. Thus, the findings support the hearing hypothe- sis, but not the vision hypothesis. An experiment can only disprove, not prove, a hypothesis. For example, one cannot conclude from the results that the hearing hypothesis is proven to be true. Perhaps the owl uses an unknown smell to strike at the mouse. One can only reject the vision hypothe- sis because it did not predict the results of the experiment correctly. Acceptance of a hypothesis is always tentative in science. The scientific community revises its understanding of phenomena, based on new data. Having ruled out one hypothesis, a biologist will devise more tests to try to rule out any remaining hypotheses. Making Inferences Scientists often draw inferences from data gathered during a field study or experiment. An inference (IN-fuhr-uhns) is a conclusion made on the basis of facts and previous knowledge rather than on direct observations. Unlike a hypothesis, an inference is not directly testable. In the owl study, it is inferred that the owl detects prey from a distance rather than by direct touch. Applying Results and Building Models As shown in Figure 1-11, scientists often apply their findings to solve practical problems. They also build models to represent or describe things. For example in 1953, James Watson and Francis Crick used cardboard balls and wire bars to build physical models of atoms in an attempt to understand the structure of DNA. Mathematical models are sets of equations that describe how dif- ferent measurable items interact in a system. The experimenter can adjust variables to better model the real-world data. CONSTRUCTING A THEORY When a set of related hypotheses is confirmed to be true many times, and it can explain a great amount of data, scientists often reclassify it as a theory. Some examples include the quantum the- ory, the cell theory, or the theory of evolution. People commonly use the word “theory” in a different way than scientists use the word. People may say “It’s just a theory” suggesting that an idea is untested, but scientists view a theory as a highly tested, generally accepted principle that explains a vast number of observations and experimental data. 6 Copyright © by Holt, Rinehart and Winston. All rights reserved. Biologists often apply their knowledge of the natural world to practical problems. Studies on the owl’s keen ability to locate sounds in space despite background noise are helping biotechnologists and bioengineers develop better solutions for people with impaired hearing, such as the people shown in this picture. FIGURE 1-11 18 CHAPTER 1 COMMUNICATING IDEAS An essential aspect of scientific research is scientists working together. Scientists often work together in research teams or sim- ply share research results with other scientists. This is done by publishing findings in scientific journals or presenting them at sci- entific meetings, as shown in Figure 1-12. Sharing information allows others working independently to verify findings or to con- tinue work on established results. For example, Roger Payne pub- lished the results of his owl experiments in a journal in 1971. Then, other biologists could repeat it for verification or use it to study the mechanisms introduced by the paper. With the growing impor- tance of science in solving societal issues, it is becoming increas- ingly vital for scientists to be able to communicate with the public at large. Publishing a Paper Scientists submit research papers to scientific journals for publica- tion. A typical research paper has four sections. First, the Introduction poses the problem and hypotheses to be investigated. Next, the Materials and Methods describe how researchers proceeded with the experiment. Third, the Results state the findings the experiment presented, and finally, the Discussion gives the significance of the experiment and future directions the scientists will take. Job Description Forensic biolo- gists are scientists who study biological materials to investigate potential crimes and other legal issues against humans and animals. Forensic scientists have knowledge in areas of biology, such as DNA and blood pattern analysis, and work in private sector and public laboratories. Focus On a Forensic Biologist As a law enforcement forensic specialist for the Texas Parks and Wildlife Department, Beverly Villarreal assists the game warden in investigations of fish and wildlife violations, such as illegal hunting and fishing. Villarreal analyzes blood and tissue samples to identify species of animals such as fish, birds, and reptiles. Her work helps game wardens as they enforce state laws regarding hunting and fishing. Most people think of forensic scientists as the glamorous crime investigators on TV, but according to Villarreal real forensic scientists “spend a great deal of time at a lab bench running analysis after analysis.” Many of the methods used in animal forensics, such as DNA sequenc- ing, are also used in human forensics. Education and Skills • High school—three years of science courses and four years of math courses. • College—bachelor of science in biol- ogy, including course work in zoology and genetics, plus experience in per- forming DNA analyses. • Skills—patience, attention to detail, and ability to use fine tools. Careers in BIOLOGY Forensic Biologist For more about careers, visit go.hrw.com and type in the keyword HM6 Careers. www.scilinks.org Topic: Scientific Investigations Keyword: HM61358 mb06se_bios03.qxd 5/18/07 10:40 AM Page 18 THE SCIENCE OF LIFE 19 1. What two principles make the scientific method a unique process? 2. Define the roles of observations and hypotheses in science. 3. Summarize the parts of a controlled experiment. 4. Summarize how we make conclusions about the results of an experiment. 5. Why is the phrase, “it’s just a theory” misleading? 6. Give another example of a conflict of interest. CRITICAL THINKING 7. Making Hypotheses On a nocturnal owl’s skull, one ear points up, and the other ear points down. Suggest a hypothesis for this observation. 8. Designing Experiments Design an experiment to establish if owls hunt by keen sight or hunt by heat seeking. 9. Calculating Information What was the average distance between the owl’s strike and the mouse if the recorded differences in this experiment were 25, 22, 19, 19, and 15? SECTION 3 REVIEW After scientists submit their papers to a scientific journal, the editors of that journal will send the paper out for peer review. In a peer review, scientists who are experts in the field anonymously read and critique that research paper. They determine if a paper pro- vides enough information so that the experiment can be duplicated and if the author used good experimental controls and reached an accurate conclusion. They also check if the paper is written clearly enough for broad understanding. Careful analysis of each other’s research by fellow scientists is essential to making scientific progress and preventing scientific dishonesty. HONESTY AND BIAS The scientific community depends on both honesty and good sci- ence. While designing new studies, experimenters must be very careful to prevent previous ideas and biases from tainting both the experimental process and the conclusions. Scientists have to keep in mind that they are always trying to disprove their favorite ideas. Scientists repeat experiments to verify previous findings. This allows for science to have a method for self-correction and it also keeps researchers honest and credible to their peers in the field. Conflict of Interest For most scientists, maintaining a good reputation for collecting and presenting valid data is more important than temporary prestige or income. So, scientists try to avoid any potential conflicts of interest. For example, a scientist who owns a biotechnology company and manufactures a drug would not be the best researcher to critically test that drug’s safety and effectiveness. To avoid this potential con- flict of interest, the scientist allows an unaffected party, such as a research group, to test the drug’s effectiveness. The threat of a potential scandal based on misleading data or conclusions is a pow- erful force in science that helps keep scientists honest and fair. Scientists present their experiments in various forms. The scientists above are presenting their work in the form of a poster at a scientific meeting. FIGURE 1-12 Copyright © by Holt, Rinehart and Winston. All rights reserved. The Internet can provide a wealth of scientific information for a report, but the information may not always be credible or accurate. You can use the methods above to check the accuracy and credibility of your sources. SCIENCE TECHNOLOGY SOCIETY SCIENCE ON THE INTERNET: A New Information Age I n the past, students research- ing a science topic would typ- ically begin their research by visiting a library to use printed reference materials, such as encyclopedias. Today, most stu- dents research topics by using a computer and searching for information on the Internet. The Internet can provide students with a wealth of infor- mation. But which Web sites have accurate information, and which Web sites do not? Checking Web Addresses Students should use the Web address, or URL, to establish the Web site’s credibility. Usually, the domain name can suggest who has published the Web site. Web sites can be pub- lished by governmental agen- cies (ends in “dot gov” or .gov), by educational institutions (ends in “dot edu” or .edu), by organizations (ends in “dot org” or .org), or by commercial businesses (ends in “dot com” or .com). Government Web sites are usually reliable. Examples of credible governmental Web sites are the National Institutes of Health (NIH) and the Food and Drug Administration (FDA). University and medical school sites are also reliable sources of information. Many organiza- tions that research and teach the public about specific diseases and conditions can also provide reliable information. Examples of such organizations are the American Cancer Society and the American Heart Association. Evaluating Web Sites The credibility of the author of the Web site should also be checked. Make sure the author is not trying to sell anything and is established in his or her field. For example, a health Web site’s author should be a med- ical professional. It is also important to check the date that the information was posted on the Web to ensure that the information is current. Also, the Web site should provide ref- erences from valid sources, such as scientific journals or govern- ment publications. Finally, the student should always double-check informa- tion between several reliable Web sites. If two or three reliable sites provide the same informa- tion, the student can feel confi- dent in using that information. Web Sites for Students The Internet Connect boxes in this textbook have all been reviewed by professionals at the National Science Teachers Association (NSTA). Students can trust that these sites are reliable sources for science- or health-related topics. REVIEW 1. Which types of Web addresses are the most reliable? 2. List four important features to evaluate when using a Web site for research. 3. Supporting Reasoned Opinions Why do you think a Web site that is advertising a product may not offer accurate information? REVIEW 20 www.scilinks.org Topic: Using the Internet Keyword: HM61589 mb06se_biosts.qxd 5/18/07 10:42 AM Page 20 TOOLS AND TECHNIQUES With proper equipment and good methods, biologists can see, manipulate, and understand the natural world in new ways. Microscopes are one of many useful tools used to unlock nature’s biological secrets. MICROSCOPES AS TOOLS Tools are objects used to improve the performance of a task. Microscopes are tools that extend human vision by making enlarged images of objects. Biologists use microscopes to study organisms, cells, cell parts, and molecules. Microscopes reveal details that otherwise might be difficult or impossible to see. Light Microscopes To see small organisms and cells, biologists typically use a light microscope, such as the one shown in Figure 1-13. A compound light microscope is a microscope that shines light through a spec- imen and has two lenses to magnify an image. To use this micro- scope, one first mounts the specimen to be viewed on a glass slide. The specimen must be thin enough for light to pass through it. For tiny pond organisms, such as the single-celled paramecium, light passing through the organism is not a problem. For thick objects, such as plant stems, biologists must cut thin slices for viewing. There are four major parts of a compound light microscope. For further description of the parts of a micro- scope, see the Appendix. 1. Eyepiece The eyepiece (ocular (AHK-yoo-luhr) lens) magnifies the image, usually 10 times. 2. Objective Lens Light passes through the specimen and then through the objective lens, which is located directly above the specimen. The objective lens enlarges the image of the specimen. Scientists sometimes use stains to make the image easier to see. 3. Stage The stage is a platform that supports a slide holding the specimen. The slide is placed over the opening in the stage of the microscope. 4. Light Source The light source is a light bulb that provides light for viewing the image. It can be either light reflected with a mirror or an incandescent light from a small lamp. SECTION 4 OBJECTIVES ● List the function of each of the major parts of a compound light microscope. ● Compare two kinds of electron microscopes. ● Describe the importance of having the SI system of measurement. ● State some examples of good laboratory practice. VOCABULARY compound light microscope eyepiece (ocular lens) objective lens stage light source magnification nosepiece resolution scanning electron microscope transmission electron microscope metric system base unit Compound light microscopes open the human eye to an interesting world including tiny pond organisms, healthy and diseased cells, and the functioning of cell parts. FIGURE 1-13 Objective lens Eyepiece (ocular lens) Stage Light THE SCIENCE OF LIFE 21 Copyright © by Holt, Rinehart and Winston. All rights reserved. 22 CHAPTER 1 Magnification and Resolution Microscopes vary in powers of magnification and resolution. Magnification is the increase of an object’s apparent size. Revolving the nosepiece, the structure that holds the set of objective lens, rotates these lenses into place above the specimen. In a typical com- pound light microscope, the most powerful objective lens produces an image up to 100 times (100) the specimen’s actual size. The degree of enlargement is called the power of magnification of the lens. The standard ocular lens magnifies a specimen 10 times (10). To compute the power of magnification of a microscope, the power of magnification of the strongest objective lens (in this case, 100) is multiplied by the power of magnification of the ocular lens (10). The result is a total power of magnification of 1000. Resolution (REZ-uh-LOO-shuhn) is the power to show details clearly in an image. The physical properties of light limit the ability of light microscopes to resolve images, as shown in Figure 1-14a. At pow- ers of magnification beyond about 2,000, the image of the speci- men becomes fuzzy. For this reason, scientists use other microscopes to view very small cells
The expression 2 + 4 1 + 2 is equal to (A) 0 (B) 1 (C) 2 (D) 4 (E) 5 2. The ones (units) digit of 542 is 2. When 542 is multiplied by 3, the ones (units) digit of the result is (A) 9 (B) 3 (C) 5 (D) 4 (E) 6 3. Some of the 1 × 1 squares in a 3 × 3 grid are shaded, as shown. What is the perimeter of the shaded region? (A) 10 (B) 14 (C) 8 (D) 18 (E) 20 4. If 3x + 4 = x + 2, the value of x is (A) 0 (B) −4 (C) −3 (D) −1 (E) −2 5. Which of the following is equal to 110% of 500? (A) 610 (B) 510 (C) 650 (D) 505 (E) 550 6. Eugene swam on Sunday, Monday and Tuesday. On Monday, he swam for 30 minutes. On Tuesday, he swam for 45 minutes. His average swim time over the three days was 34 minutes. For how many minutes did he swim on Sunday? (A) 20 (B) 25 (C) 27 (D) 32 (E) 37.5 7. For which of the following values of x is x 3 < x2 ? (A) x = 5 3 (B) x = 3 4 (C) x = 1 (D) x = 3 2 (E) x = 2112 years, Janice will be 8 times as old as she was 2 years ago. How old is Janice now? (A) 4 (B) 8 (C) 10 (D) 2 (E) 6 10. In the diagram, pentagon T P SRQ is constructed from equilateral 4 P T Q and square P QRS. The measure of ∠ST R is equal to (A) 10◦ (B) 15◦ (C) 20◦ (D) 30◦ (E) 45◦ Q P R S T Part B: Each correct answer is worth 6. 11. In the diagram, which of the following points is at a different distance from P than the rest of the points? (A) A (B) B (C) C (D) D (E) E y A x 2 2 4 4 6 8 6 8 B C D E P 12. If x = 2 and y = x 2 − 5 and z = y 2 − 5, then z equals (A) −6 (B) −8 (C) 4 (D) 76 (E) −4 13. In the diagram, P QR is a straight line segment. If x + y = 76, what is the value of x? (A) 28 (B) 30 (C) 35 (D) 36 (E) 38 x° x° x° y° y° P Q R 14. The line with equation y = 2x − 6 is reflected in the y-axis. What is the x-intercept of the resulting line? (A) −12 (B) 6 (C) −6 (D) −3 (E) 0 15. Amy bought and then sold 15n avocados, for some positive integer n. She made a profit of $100. (Her profit is the difference between the total amount that she earned by selling the avocados and the total amount that she spent in buying the avocados.) She paid $2 for every 3 avocados. She sold every 5 avocados for $4. What is the value of n? (A) 100 (B) 20 (C) 50 (D) 30 (E) 8 16. If 3x = 5, the value of 3x+2 is (A) 10 (B) 25 (C) 2187 (D) 14 (E) 45
The Revolt of the Northern Earls (1569) Most people in the North remained loyal to the Catholic noble families who controlled the north and their Catholic faith. When Elizabeth came to power, she promoted ‘new men’ (Protestants) from the gentry and the powerful Catholic nobles lost their power and influence. This led them to organise the most serious rebellion of Elizabeth’s reign in 1569. Why did the Northern Earl’s revolt? The Earls had lost their power when Elizabeth became Queen (and wanted it back). They wanted Catholicism restored in England (and felt that ordinary Catholics would support it). Elizabeth was refusing to marry or to name an heir, causing uncertainty about England’s future. Mary Queen of Scots (if freed from prison) could replace Elizabeth and solve all these problems Who were the key players in the Revolt? Earl of Northumberland • A Catholic who had held an important position under Mary I. • He lost a lot of influence under Elizabeth (as she favoured Protestant gentry) • Elizabeth also took the rights to a valuable copper mine found on his lands Earl of Westmorland • From a rich Catholic family in the north Also the Duke of Norfolk’s brother in law Duke of Norfolk • England’s most senior Protestant noble, but he had very close links to old northern Catholic families, & was sympathetic to them & greedy for power. • He hated William Cecil & Robert Dudley, Earl of Leicester (Elizabeth’s favourite) who were Protestant and from the gentry • He planned to marry Mary QS, but later backed down and urged the earls to call off the rebellion. Mary also supported the plan to marry him What role did religion play? (7/10 – but only because it was linked to power) • Most northerners held onto their Catholic beliefs & although Elizabeth didn’t persecute them, they knew that she wanted their religion to gradually die out, so they supported the revolt. • In 1561 Elizabeth hired a strict Protestant as archbishop of Durham to promote Protestantism in the north, but he was unpopular & turned many northerners against the Protestant religion. What role did politics/power play? (9/10 – this was the most important cause of the revolt) • The Northern Earls lost a lot of their power/influence (even jobs/money under Elizabeth) • Northumberland was jealous of new Protestant families being given top jobs in the North • William Cecil & Robert Dudley were not from ancient noble families, but were very close to the Queen, so the northern Earls resented them getting top jobs in her Government • Elizabeth also confiscated large areas of land & the profits from their copper mines • It is possible, that had Elizabeth allowed the Catholic Northern Earls to keep their jobs, money and influence at court, they may have ‘tolerated’ her as a Protestant Queen (greedy/selfish). What role did Mary Queen of Scots and the Succession play? • Elizabeth was refusing to name an heir and it was becoming clear that she would not marry • If Mary Queen of Scots married the Duke of Norfolk, England would have an heir and England would be Catholic again. The country would be stable without people competing for power. • However, some of Elizabeth’s courtiers got worried that it might not work and that it might lead to charges of treason (punishable by death) • So by September 1569, Robert Dudley (Earl of Leicester) decided to tell Elizabeth about the plot. By this time it was much more serious than simply marrying Norfolk to Mary. • Mary QS had secretly asked Spain to send troops to help the rebellion & overthrow Elizabeth Plan for the Revolt of the Northern Earls (1569) • The Earls of Northumberland & Westmorland will raise rebel troops from their lands in the north and take control of Durham. • The rebels will then march south towards London to join with the Duke of Norfolk • 1000s of Spanish troops will land in England to support the rebel forces • The Duke of Norfolk & rebel forces will seize control of Government & overthrow Elizabeth • Mary Queen of Scots is to be freed, ready to marry the Duke of Norfolk Key Events of the Revolt • Once Elizabeth knew of the plot, Norfolk was arrested and sent to the Tower of London • The Northern Earls were worried they would be executed for their involvement and in a desperate attempt to avoid punishment, pushed ahead with the revolt • They raised an army of ordinary Catholics and took control of Durham cathedral • Catholic mass was celebrated across the north for 2 weeks. • They then headed south, to try and free Mary • Mary QSs was moved south to Coventry on the orders of Elizabeth, so she couldn’t escape • The rebellion failed as Spanish troops never arrived • Elizabeth’s friend (Earl of Sussex) had raised an army of 7,000 men to defend her throne. Results: • The rebellion was a serious threat to Elizabeth • She executed 450 rebels in the north • Northumberland was executed in 1572 & his head was put on a spike on the city gate • The Privy Council called for the Duke of Norfolk’s execution too, but Elizabeth released him. • Mary Queen of Scots was kept in prison for the next 14 years. • The failed plot also led the Pope to take action against Elizabeth • In 1570 he excommunicated Elizabeth from the Catholic Church • He also issued a Papal Bull (order) calling on all loyal Catholics to overthrow her hoping it would encourage another rebellion. • In 1571 Elizabeth called parliament to pass an Act making it treason to claim that she was not the rightful Queen and to bring in/print papal bulls in England. The Significance of the Revolt of the Northern Earls • It was the first and most serious rebellion by English Catholics against Elizabeth • Treason laws were made much harsher • It ended the influence of the powerful Catholic Earls in the North • It led to harsher treatment of Catholics, e.g. 1572 Elizabeth sent the Earl of Huntingdon (strict Protestant) to the north to carry out laws against Catholics (and suppress Catholicism). • Although Elizabeth’s brutal revenge on the rebels show how serious a threat it was, most Catholics in the north stayed loyal, but the Pope’s Papal Bull now put their loyalty in doubt There was little support for the revolt among the rest of the Catholic nobility and ordinary people. When faced with a choice between Elizabeth and their religion, most Catholics chose to support the Queen. 1569, was the last time English Catholics tried to remove Elizabeth by force. The future plots against her were always uncovered by Cecil & Walsingham, before they had a chance to get any public support. Despite this, the Northern Revolt & Papal Bull changed Elizabeth’s attitude towards Catholics who were now seen as potential traitors. From 1570, Elizabeth became less tolerant of recusants (people refusing to attend her church) & took increasingly tough measures against Catholics. The Ridolfi, Throckmorton & Babington plots • In the 1870s-80s, there were 3 Catholic plots to assassinate Elizabeth & replace her with Mary. • The plots were supported by France, Spain, the Pope and some Catholic nobles. • They reinforced the form Mary & from Catholics at home and abroad. Also the threat from Spain. The Ridolfi Plot (1571) • Ridolfi was an Italian banker living in England and a spy for the Pope. • He organised a plot to murder Eliz, marry Mary QS to the Duke of Norfolk & make her Queen. • The Pope & King Philip supported the plot & Philip told the Duke of Alba in the Netherlands to prepare 10,000 troops (but to only invade AFTER the English had overthrown Elizabeth). • The plot failed because Sir William Cecil intercepted coded letters & Norfolk was executed. • Mary was kept under closer watch. • Ridolfi was abroad when the plot was discovered and never returned to England. 1574: Catholic Priests and Priest Holes • From 1574 Catholic priests were smuggled into England to keep the religion alive. • They stayed with rich Catholic families, so Catholic families were kept under surveillance. • Catholic homes were raided – to find ‘priest holes’ where Catholic priests were hiding. • Catholic priests who were found could be hung, drawn and quartered (although not all were) • In 1581, Parliament also passed 2 new tougher laws against Catholics: • Recusants would be fined £20 (which would bankrupt most families) • Trying to convert people to Catholicism was now treason (punishable by death) The Throckmorton Plot (1583) • It aimed to assassinate Elizabeth and replace her with Mary. The French Duke of Guise (Mary’s cousin) would invade England with an army, funded by King Philip (Pope also supported it). • An Englishman, Throckmorton carried messages between Mary & Catholic plotters abroad. • Sir Walsingham (Secretary of State) uncovered the plot after his agents found the plans for the plot in Throckmorton’s house. Throckmorton confessed under torture and was executed. Significance: • The plots showed that Mary’s presence in England posed a serious threat • It also showed that France & Spain were a serious threat (& could invade) • Throckmorton’s papers showed a list of Catholic supporters in England, so the threat from English Catholics was also real • 1,000s of Catholics were imprisoned or kept under surveillance/house arrest • In 1585 another Act was passed to make helping Catholic priests punishable by death. • The Bond of Association was signed by the English nobles & gentry & forced them to promise to execute anyone who tried to overthrow the Queen. Weaknesses of the Plots The plots lacked public support & were uncovered by informers & spies before they had the chance to work King Philip was reluctant to destroy his alliance with Elizabeth (France was still a bigger rival) so is support for the plots was half-hearted, he rarely followed through on his promises to help the plotters or send an army The Babington Plot (1586) In 1586, Walsingham used his spy network to PROVE that Mary supported the Babington plot. His evidence persuaded Elizabeth to put Mary on trial & execute her for treason. • This was a plot to murder Elizabeth and put Mary on the throne • France would invade England with 60,000 men and Spain would also send an army • Babington was passing coded letters between Mary & her supporters in England & Europe. • But all of her letters were being intercepted and read by Walsingham. • Walsingham used his spies to follow every stage of the plot & had the letters decoded • One of Mary’s letters approved plans to murder the Queen and free Mary from prison • They also contained the names of 6 Catholics who planned to kill Elizabeth • They were arrested, hung, drawn and quartered for treason. • Mary had been implicated in plots before, but Elizabeth was always reluctant to execute her • But the proof found by Walsingham finally persuaded her to put Mary on trial • In October 1586, Mary was found guilty & was sentenced to death • But Elizabeth still hesitated, and did not sign the death warrant until February 1587. Significance 1) This plot was very significant because by 1585 England was effectively at war with Spain since Elizabeth had sent her army to help the Dutch Protestants fight the Spanish 2) This meant that Elizabeth’ situation was more dangerous than during previous plots. 3) Elizabeth’s government also became more determined to crush Catholicism 4) 1000s of recusants were arrested & 31 priests were executed 5) Mary’s execution removed the Catholic threat at home 6) English Catholics had no one to rally around, & lost hope of overthrowing Elizabeth 7) But Mary’s death increased the threat of a foreign invasion as England was at war with Spain and King Philip had been preparing an attack on England since 1585 8) Mary’s death made Philip even more determined to invade, Mary had left her claim to the English throne to King Philip upon her death Why was Mary Queen of Scots finally executed? 1 • A new Act in 1585 stated that in the event of Elizabeth’s assassination, Mary could be executed as long as she had been proved guilty & Walsingham had provided hard proof. 2 • Another reason was that by 1587, it was clear that Philip was planning to invade England • There were rumours that Spanish ships had landed in Wales & that Mary had escaped. This convinced Elizabeth that Mary had to be executed if she wanted to keep her throne. Walsingham’s Spy Network: • Walsingham (Secretary of State from 1573) had a network of spies all over England & abroad. He had spies in every English town, some were normal people paid to spy on neighbours. • He also had agents and spies in Spain, France, Germany and Italy • He hired mathematicians to crack written codes and people to open/seal letters secretly • He also pressured captured Catholic priests to spy on others for him in return for a pardon. • He used double agents to infiltrate Catholic networks - to help him discover traitors • But he only used torture against Catholic priests caught in England in the most serious cases • But 130 priests and 60 of their supporters were still executed during Elizabeth’s reign. Why did Relations with Spain get worse (1569-1588) • England had tried to stay on good terms with Spain, because Eliz wanted to avoid an expensive war that could lead to her being overthrown (English Catholics could support it) • But by the 1570s, Elizabeth wanted to have an empire of her own. • She also needed to make more money to defend her country and throne (by improving trade) • This religious, political and economic rivalry led to growing tensions between England & Spain Political and Religious Rivalry 1) Land abroad, gave countries wealth/power. By the 1580s, Eliz wanted an empire to rival Spain’s (especially as Spain had supported the Catholic plots against Eliz – even if it was half-hearted) 2) Religion was another cause of conflict. Philip opposed Elizabeth’s religious settlement 1559 3) Luckily for Elizabeth, in the 1550s Spain & France were competing to be the greatest European power and both wanted England as an ally against the other. 4) But from 1567, Spanish ships were sailing to the Netherlands with money for the Alba’s army 5) This alarmed English Protestants and Elizabeth’s Privy Council who put more and more pressure on her to send an army to help the Dutch Protestant rebels (in the Netherlands). Economic (commercial) Rivalry: The New World, privateers and Sir Francis Drake • Under Elizabeth, English merchants wanted to make big profits in the New World (Americas). • However, trading in the New World was difficult because of Spain’s power 1) Spain controlled most of the New World where there were huge profits to be made and anyone who wanted to trade there needed a licence from Spain (which it would not give): 2) But the Americas had valuable crops like tobacco, sugar, and also silver and gold 3) Elizabeth secretly encouraged privateers to trade illegally & raid Spanish ports & ships 4) At first Elizabeth denied responsibility for their actions, which delaye war with Spain Sir Francis Drake: Elizabeth sends Drake to rob Spanish colonies and ships (which infuriates Spain) 1) Spain’s support for the Ridolfi plot (1571) made her more willing to support Drake • In 1572 Eliz hired Drake to sail to the New World & steal £40,000 of Spanish silver • In 1577 she sent Drake back again with a secret mission to rob Spain’s colonies/ships • Drake brought back £400,000 of Spanish treasure & claimed an area of California in Elizabeth’s name (New Albion). He gave a lot of this money to Elizabeth • He boosted England’s finances at a time of growing concern over Spain’s threat • He became famous as the first Englishman to circumnavigate the globe. • Eliz knighted Drake as a reward, which infuriated Philip (as he saw Drake as a pirate) • Drake’s actions & his claim to California made it clear that England did not accept Spain’s domination of the New World. Elizabeth’s Support for the Dutch Rebels led to War with Spain (1585-88) • By the 1580s, tension between England & Spain had reached boiling point • At first, Eliz refused to send her army to help the Dutch rebels, because she wanted to avoid a war with Spain. So she tried to get the Spanish to leave the Netherlands in other INDIRECT ways: 1) By allowing Drake (& other English privateers) to attack and rob Spanish ships and colonies 2) By encouraging others (the French heir/mercenaries) to fight the Spanish in the Netherlands • In the 1570s, Elizabeth promised to marry the heir to the French throne (the Duke of Alencon) so that he would take an army to fight the Spanish in the Netherlands The Spanish Fury (1576) and the Pacification of Ghent (1576) • By 1576, the Spanish Govt in the Netherlands was bankrupt (the war was expensive) • After months without pay, Spain’s soldiers violently robbed Dutch towns in the “Spanish Fury” Spanish troops rebelling and robbing cities in the Netherlands in 1576. This united the Dutch Protestants & Catholics against Spain. They drew up the ‘Pacification of Ghent’ (demanding that): • Spanish troops leave the Netherlands • Spain allows the Dutch to rule themselves • The persecution of Dutch Protestants stops What did Elizabeth do? • Elizabeth sent £100,000 to help the Dutch rebels • In 1577 King Philip’s brother, Don Juan agreed to the rebels demands (but this was a trick) as just 6 months later Philip sent an even bigger army to attack the Dutch. • Elizabeth then hired a mercenary army of 6000 English & Scottish volunteers to help the Dutch. • But her plan backfired because the mercenaries destroyed Dutch Catholic churches, which caused the Catholics to make peace with Spain. • In 1578, her Privy Council urged Eliz to send her official army to help the Dutch, but she refused. The Dutch were disappointed & turned to France for help. The French Duke of Alencon arrived with an army to fight the Spanish, but by 1579 Spain had taken control again. • In 1580 Spain got even stronger after Philip won control of Portugal & its empire. • So Elizabeth gave the Duke of Alencon £70,000 to help him fight the Spanish • In 1582, Alencon took his army the Netherlands but failed to defeat Spain. • Elizabeth’s foreign policy in the Netherlands had failed & she had only managed to annoy Spain 1585: Why did Eliz finally decide to send her army to the Netherlands? (she lost her 2 main allies) • 1584 the Duke of Alencon died (so he could no longer fight the Spanish in the Netherlands) • 1 month later, William of Orange, the leader of the Dutch Protestant rebels was assassinated. • In 1585, Spain signed the Treaty of Joinville with France, agreeing to stamp out Protestantism in France/Europe meaning France & Spain were now allies against Protestantism • Elizabeth now felt she had no choice but to send her official army to the Netherlands • She signed the Treaty of Nonsuch with the Dutch rebels which promised them military help 1585: Robert Dudley’s campaign in the Netherlands was unsuccessful She sent 7,400 man army to the Netherlands led by Dudley. But he accepted the title of ‘Governor General’. Eliz was angry as it suggested that she had deposed King Philip so she told Dudley to resign this position. His army was defeated by the bigger Spanish Army as Eliz had not provided him with enough money to win. In 1587 Dudley resigned and returned to England. At the same time, Eliz had sent Drake to raid Spanish colonies in the New World to disrupt King Philip’s flow of money. Philip was furious and told the Pope he planned to invade England at the end of 1585. Drake singes the King of Spain’s beard 1587 • In 1587 Elizabeth ordered Drake to attack Spain’s most important port Cadiz • He destroyed 30 ships in 3 days – known as the ‘Singeing of the King of Spain’s Beard’ • He also stole lots of wood, meaning the Armada did not have quality barrels for food/water • Drake’s disruption delayed the Armada by a year (& meant that its food rotted in 1588). • This bought England more time to prepare for war. The Spanish Armada (1588) The Plan • By 1588, the Spanish Armada was ready to invade England • It had 130 ships with 8000 sailors & 18,000 soldiers • The Duke of Medina Sidonia would lead the Armada, but he had little experience at sea and didn’t want the job • The Armada would collect Parma’s army from France & sail to England under the protection of the Armada’s warships • Parma would march to London to depose Elizabeth & impose a Catholic government in England. 1) The Armada reached the English Channel The Armada set out in May 1588, but was delayed for a few weeks by bad weather In July the Armada was near England & signal fires were lit to warn Elizabeth English ships set sail to meet the Armada The Armada sailed up the channel in a crescent (half moon) formation, to use the large armed galleons to protect the weaker supply and army ships The English navy carried out a few minor raids, but did not inflict much damage Only 2 Spanish ships were lost (by accident) 2) The English attack the Spanish at Calais (with fire ships) and at Gravelines The Armada sailed up the English channel & anchored at Calais to wait for Parma’s army But Parma’s men didn't reach the coast in time (news had reached them too late) At midnight, the English sent 8 fireships into the Spanish ships causing panic They cut their anchors, broke formation & headed for the open sea (without Parma) The Spanish ships sailed to Gravelines, but bad weather stopped them returning to Calais The English attacked and the battle lasted many hours (5 Spanish ships were sunk) The rest were forced to sail away from France towards Scotland The English ships followed them to make sure they didn’t come back to collect Parma’s army 3) The Armada’s Journey back to Spain around Ireland was a disaster The Spanish called off the attack and returned to Spain around Scotland & Ireland Bad storms sank many ships and wrecked more on the Irish coast Many sailors died from starvation & disease – less than half the men made it back to Spain How did England defeat the Spanish Armada? !) Faster Ships • Years before the battle, England had started building smaller, faster ships (galleons) that could fire canon balls quicker & further than Spanish ships • Spanish ships were huge and slow to change direction. 2) Bad Planning & Communication (Spanish) • Philip’s plan to join with the Duke of Parma’s army in France was risky. • Parma had lots of small ships which took 48 hours to load, man and set sail. • It took too long (a week) for word to reach Parma that Medina was in the English Channel, by which time Medina had set sail to Calais. • Parma was not ready to set sail & the English were already ready to attack (leaving Medina with very little back up when anchored in France). 2) English Tactics were more effective • Spanish ships aimed to come alongside the English ones, jump on board & fight the enemy. But the English ships were faster & kept a safe distance. • They chased the Armada down the Channel, with heavy cannon fire, which forced the Spanish to arrive in France before Parma’s army was ready • As the Armada was waiting, the English sent fireships into the Spanish fleet. • This caused the Armada to panic, cut their anchors & sail away to the north • When the Spanish ships regrouped, the English attacked them in the Battle of Gravelines & the Armada was forced to sail north, chased by faster ships. 5) Bad Weather • Strong winds made it impossible for the Armada to return & pick up Parma’s army and storms wrecked or sunk Spanish ships as they tried to return home along the Scottish-Irish coasts. 2) Spanish Supplies • The Armada was not well supplied with food/weapons. Drake’s attack on Cadiz port in 1587 had destroyed food barrels. Delays in setting sail meant that by the time the English attacked the Armada it had been at sea for 10 weeks and had rotting food. 1000s died from starvation/disease. The consequences of the English victory? • Victory over the Spanish Armada gave Elizabeth a great propaganda victory • A new portrait was made, and a medal was made to commemorate her victory, it said “God blew and they were scattered”. • Elizabeth claimed that God was on the side of Protestantism • This led to a feeling of English pride and encouraged the Dutch rebels to renew their fight against the Spanish • The defeat of the Armada showed the strength of the English navy and gave England the confidence to trade and explore more widely at sea • Although Philip did not give up and continued the war for the rest of Elizabeth’s reign, the defeat had cost Spain dearly, both financially and in terms of its power • The Armada marked the start of a long decline in Spain’s power and fortunes. • English ships were sent on voyages of discovery and set up valuable new trade routes • By the end of Elizabeth’s reign, the navy was also trying to set up a new colony in Virginia • The English victory boosted Elizabeth’s popularity & strengthened the Protestant cause