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Lesson 1.1 through 1.8
Quiz by David Neuman
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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.Here’s a **quiz on Lesson 1: Introduction to Analog Communication (Unit 8)** based on your file 👇 --- # 🧠 **Quiz – Lesson 1 (Analog Communication)** **Marks:** 20 --- ## ✍️ **Part 1: Choose the correct answer (8 marks)** 1. A signal is: a) A device b) A physical quantity that carries information c) A type of wire d) A computer 2. A continuous signal is defined over: a) Discrete values b) Infinite real values c) Only integers d) Binary values 3. Digital signals have: a) Infinite values b) Two values (0 and 1) c) Random values d) Analog values 4. Sampling is used to: a) Increase noise b) Convert analog to digital c) Amplify signals d) Reduce bandwidth 5. A deterministic signal: a) Cannot be predicted b) Has known values c) Is always random d) Has no pattern 6. Even signal satisfies: a) x(t) = -x(-t) b) x(t) = x(-t) c) x(t) = 0 d) x(t) ≠ x(-t) 7. Periodic signal repeats after: a) Time T b) Infinite time c) No time d) Random time 8. A system is: a) A signal only b) Input only c) Takes input and gives output d) A wire --- ## ✍️ **Part 2: Complete (6 marks)** 1. A signal can be represented as __________. 2. Continuous signals are defined over __________ values. 3. Digital signals take values like __________ and __________. 4. A random signal cannot be __________ easily. 5. Odd signal satisfies __________. 6. A periodic signal repeats every __________. --- ## ✍️ **Part 3: True or False (6 marks)** 1. Analog signals are continuous. ( ) 2. Digital signals can take infinite values. ( ) 3. Sampling converts analog to digital signal. ( ) 4. Deterministic signals are predictable. ( ) 5. Odd signals pass through origin. ( ) 6. Aperiodic signals repeat over time. ( ) --- ## 🎯 **Bonus Question (Optional)** Give one example of: * Analog signal * Digital signal -Lesson 2: USES OF SOIL PRETEST Color the pictures that show how we can use soil. SOMETHING TO READ Soil is very useful to us. We can use it in many ways. Let us find out the different uses of soil one by one. Uses of soil a. Sand is used in making our houses. It is also used in making hollow blocks. OBJECTIVES: - Enumerates things that we can with soil - Demonstrates ways for making play things out of soil SCIENCE 2 – MODULE 7 SEIBO COLLEGE 9 b. Soil is made up of minerals, nutrients, water and air that support growing plants. It also keeps the plant’s roots on the ground. c. Animals like earthworm and ants lives in the soil. They create tunnels in it to allow air and water to pass through it. d. Clay soil is used in making pots and vases. We can play with it. We can make different objects that we can use to play with. e. When we are at the shore of the beach, we can play with sand and build sand castles. SOMETHING TO DO ACTIVITY 1 Creating Things I Like Objective: In this activity, children will learn that they can create things out of clay. What you need: SCIENCE 2 – MODULE 7 SEIBO COLLEGE 10 2 bars of clay (any color) 1 pc. 1/8 illustration board or any hard board What to do: 1. Using the bars of clay, create things that you like then place them on the illustration board. 2. Show your work to your facilitator. Observation: 1. Is it easy for you to create things that you like out of clay? __________________________________________________ 2. What is the texture of the clay? ________________________ 3. Did you have fun dong this activity? _____________________ Conclusion: I therefore conclude that _Make the following info into a quiz: Questions (20) Show answers 1 - Quiz I ________ this would be difficult. 20sec 2 - Quiz She ______ a good lesson. 20sec 3 - Quiz Even ________ we are tired, we can do this. 20sec 4 - Quiz That's a ______ question. I don't know the answer. 20sec 5 - Quiz Have a _______ look through your exam paper before you hand it in. 20sec 6 - Quiz I received 10 calls from your mom _____ the day - just call her back! 20sec 7 - Quiz He wore shorts even ______ it was cold. 20sec 8 - Quiz The elephant's skin is very ______. 20sec 9 - Quiz What a great _____ to bring her flowers! 20sec 10 - Quiz We drove _______ a storm. 20sec 11 - Quiz We made it ______ the first semester. 20sec 12 - Quiz I need a _____ report on the latest trends in online shopping. 20sec 13 - Quiz I liked her outfit, I didn't like her accessories _____. 20sec 14 - Quiz We were _____ a lot by the pandemic. 20sec 15 - Quiz I have never _____ about relocating. 20sec 16 - Quiz She looked ______ the magazine. 20sec 17 - Quiz We went ________ the tunnel. 20sec 18 - Quiz That was a ______ hike! I am tired. 20sec 19 - Quiz The house was renovated _____ - everything is new, even windows. 20sec 20 - Quiz Wait a moment, I'll put you _____ to the sales department.LESSON 4. Cellular Respiration • Define cellular respiration • Identify the stages of clan respiration You have just learned how the energy from the sun is captured, processed, and stored in the form of glucose. Cellular respiration, another important life process, is the means by which cells release the stored energy in glucose to make adenosine triphosphate (ATP). The primary goal of this life process is to convert stored energy into usable form, such as ATP, for the cells to carry out their functions. Cellular respiration involves several chemical reactions. The reactions can be summed up in the following equation: C6 H12 O6 + 602 ----- 6 CO₂ +6H₂O + ATP Glucose oxygen carbon dioxide water energy Aerobic respiration reactions, or cellular respiration that takes place in the presence of oxygen, can be grouped into three stages glycolysis, Krebs cycle, and electron transport chain (ETC). Stage 1: Glycolysis Glycolysis is the process that breaks down one molecule of 6-C glucose into 3-C pyruvates or pyruvic acids. It also releases four molecules of ATP. This process occurs in the cytoplasm of the cell. The following is the step-by-step process of glycolysis. Take note that several enzymes are involved in this process. 1. The first step of glycolysis requires energy. It can only proceed when the two ATP molecules donate energy to the glucose by transferring a phosphate group with the help of an enzyme, producing glucose 6-phosphate 2. Then, a specific enzyme promotes the rearrangement of the atoms, producing the fructose 6-phosphate. 3. The action of the enzyme in step 2 promotes the transfer of a phosphate group from another ATP molecule, forming fructose 1,6-bisphosphate. 4. The resulting fructose 1,6-bisphosphate molecules, with the help of another enzyme, splits into two molecules, each with three carbon backbones. These two sugars are dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. 5. Another important enzyme then rapidly interconverts the molecules of dihydro-xyacetone phosphate and glyceraldehyde 3-phosphate. This produces two molecules of glyceraldehyde 3-phosphate or 3-phosphoglyceraldehyde (PGAL) 6. The succeeding step involves another enzyme-mediated action. The hydrogen (H) from PGAL is transferred to the oxidizing agent, nicotinamide adenine dinucleotide (NAD), which forms NADH. A phosphate (P) is also added from the cytosol of the cell to oxidize the two molecules of PGAL, forming two 1.3-bisphosphoglycerate. 7. A phosphate (P) from 1,3-biphosphoglycerate is transferred to ADP to form ATP. This happens for each of the two 1,3-bisphosphoglycerate. resulting to a yield of two ATP and two 3-phosphoglycerate molecules. 8. A phosphate is transferred from 3-phosphoglycerate molecules from the third carbon to the second carbon, forming 2-phosphoglycerate molecules A hydrogen atom and a hydroxyl ((OH) group is released, which then combines to form water (H2O). The removal of H2O from 2-phosphoglycerate results in the formation of 2- phosphoglycerate molecules. 9. A hydrogen atom and a hydroxyl ((OH) group is released, which then combines to form water (H2O). The removal of H2O from 2-phosphoglycerate results in the formation of two phosphoenolpyruvic acid (PEP) 10. Phosphate (P) from PEP is transferred to ADP (and forms ATP) and the final product, pyruvic acid. This reaction yields two molecules of pyruvic acid and two ATP molecules In summary, a single glucose molecule that undergoes the process of glycolysis produces two molecules of pyruvic acid, four molecules of ATP, two molecules of NADEL and two molecules of H.O. However, only two molecules of ATP are counted as net products since two molecules of ATP are spent throughout the process. Stage II: Krebs Cycle The Krebs cycle, named after its proponent Sir Hans Adolf Krebs, is a cyclical series of enzyme-controlled reactions. This stage of cellular respiration occurs in the matrix of the mitochondria. It is sometimes. called the citric acid cycle (CAC) since it produces citric acid. Citric acid contains three carboxyl (COOH) groups; hence, it is also called the tricarboxylic acid cycle (TCA). This requires the pyruvic acids produced during glycolysis. The main function of this cycle is to produce high-energy-yielding molecules, namely, NADH and flavin adenine dinucleotide (FADH) that will later on be used in the electron transport chain reaction. Figure 6-7. Summary of glycolysis and corresponding products in each reaction presented (See Appendix F on page 285 for an enlarged and complete version of the image.) An initial process is needed for the Krebs cycle to begin. As a pyruvate molecule from glycolysis enters the mitochondrion, it undergoes an important preliminary ate to form acetyl-CoA reaction. Coenzyme-A (COA) combines with pyruvate help of an enzymatic complex. This conversion also produces CO, and NADH. The Krebs cycle is summarized as follows. Take note that several enzymes are involved in this process. 1. The Krebs cycle technically begins when the acetyl-CoA combines with oxaloacetic acid (OAA), a 4-C molecule, to produce citric acid, a 6-C molecule. 2. With the aid of an enzyme, the citric acid now goes through a series of reactions that releases energy. Water molecule is removed from the citric acid and is returned in a different location. The-OH group is repositioned, forming the molecule isocitrate. 3. Isocitrate is then oxidized, forming the a-ketoglutarate, a 5-C molecule. The byproducts of this reaction are NADH and CO, 4 The a-ketoglutarate loses its CO, and a coenzyme-A is added in its place. The decarboxylation occurs with the help of NAD, which then becomes NADH. The resulting molecule is called succinyl-CoA. 5. Succinyl-CoA is converted into succinate. Also in this reaction, a molecule of guanosine triphosphate (GTP) is synthesized. The GTP molecule has similar structure and energy properties to that of ATP and is used by cells the same way. The free phosphate group attacks the succinyl-CoA molecule, which detaches the COA. Then, phosphate is attached to GDP to come up with GTP, similar to the process that occur in ATP synthesis (from ADP to ATP). 6. Two hydrogens are removed from succinate, A molecule of flavin adenine dinucleotide (FAD), a coenzyme similar to NAD, is reduced to FADH, as it takes the hydrogens from the succinate. This reaction produces the fumarate. 7. Fumarate is then converted into malate as the addition of a water molecule is catalyzed. The final reaction is the regeneration of oxaloacetate. The resulting byproduct of this regeneration is NADH Recall that two pyruvate molecules were produced during glycolysis, causing the Krebs cycle to turn twice. Each tuts produces three molecules of NADH, single ATH one FADIH, and the by-product CO, which is exhaled. Stage III: Electron Transport Chain The electron transport chain (ETC) is a series of photon pumps on the inner membrane of the mitochondrion. Electron transport is the last stage of the cellular respiration. In this stage, the energy from NADH and FADH, from the Krebs cycle is transferred to ADP to produce ATP. This process is generally known as oxidative phosphorylation. This energy coupling mechanism in the cell was revealed by the work of Peter stored energy in the form of proton (1) gradient to phosphorylate (add phosphate) ADP and produce ATP. The pumping of hydrogen sons across the inner membrane creates higher concentration ions in the inner membrane than on the outside of the membrane. This chemiosmotic gradient causes the ions to flow back across the membrane where the concentration of ions is lower. ATP synthase lined in the matrix serve as a channel protein, helping the ions to move across the membrane. The chemiosmotic gradient powers the phosphorylation of ADP to ATP, which also occurs in the ATP synthase. After passing through the ETC, the oxygen, being the final hydrogen acceptor, combines with two electrons and two protons, forming a water molecule. Water is a by-product of cellular respiration and is excreted. MINI TEST 6-3 1. Which energy-releasing pathway yields the most ATF in each glucose molecule? 2. Briefly describe the two stages of aerobic respiration that follow glycolysis: (a) Krebs cycle (b) Electron transport chain Anaerobic Respiration Most cells carry out arrobic respiration when oxygen is present. Aerobic respiration is an efficient process that yields a lot of ATP. However, many organisms thrive in mud, marshes, animal gut, canned goods, sewage treatment pond, and deep oceans where oxygen is scarce. Organisms that can live without oxygen are called anaerobes. Cellular respiration that proceeds without the presence of oxygen is called anaerobic respiration. In the event that the oxygen supply becomes low, aerobic cells also perform fermentation and lactic acid fermentation anaerobic pathways. There are two common anaerobic pathways in these cells, alcoholic fermentation and lactic acid fermentation. In alcoholic fermentation, ethyl alcohol and carbon dioxide are produced by some cells using the pyruvate from glycolysis. Each pyruvate molecule is rearranged into acetaldehyde and carbon dioxide, which is eventually released. NADII gives up electrons to acetaldehyde to form ethanol Fermentation is widely used in the industry. Yeast, a fungus used in making bread. can undergo anaerobic respiration. Bakers aux sugar, flour, water, and yeast to form the bread dough. The dough rises due to the carbon dioxide and alcohol released by the yeast cells trapped in air bubbles. Beer and wine manufacturers, we yeast to ferment the sugars in wheat and grape juice, forming alcoholic beverages such as beer and wine. In some cells, glycolysis produces two pyruvates, two NADH molecules, and two ATP molecules. Pyruvate itself becomes the final acceptor of the electrons from the NADH that produces the final product: lactate. Oftentimes, this product is called lactic acid. Human skeletal muscles can carry out fermentation when the blood cannot supply the cells with adequate oxygen during strenuous activities. When lactic acid builds up in the muscles, fatigue, burning sensation, and cramps result. Lactic acid will continue to build up until there is adequate supply of oxygen. Lactic acid is then converted back into pyruvate in the liver. Muscles also restore normal functions. Have you ever wondered why milk or cream turns sour after some time? Bacterial cells that undergo fermentation are responsible in producing lactate that turns the milk sour. These bacteria are used in manufacturing yogurt and sour milk products. Fermentation pathways do not breakdown and utilize the glucose completely. ATP is no longer produced beyond the process of glycolysis. Thus, energy produced is just enough for some single-celled organisms, or the energy can only be used by multicellular organisms for a short period.LESSON 2 Early Theories on the Origin of Life • Identify the different theories on the origin of life. • Describe each theory and determine their differences What are the characteristics of living things? Before learning about the history of Earth based on geological evidence, early scientists explored the possibilities of how the first life-form existed. There are several theories about the origin of life. Theory of Catastrophism The theory of catastrophism supported by French scientists Georges Cuvier (1769-1832) and Alcide Dessalines d'Orbigny (1802-1857), is said to be a modification of the creation story of the Bible. It states that there have been several living creations from God, each encountered a catastrophe that completely destroyed them. Each new creation consisted of new life-forms, which happen to be different from the previous ones. Theory of Abiogenesis The theory of abiogenesis, or the spontaneous generation theory, states that living things were naturally created from nonliving things such as simple organic compounds. The theory supposes that abiogenesis occurred between 3.8 and 4 Gya. The experiment performed by Stanley Miller in 1953 gave way to many speculations and studies on how life on Earth really began. His research involved a simulation of the possible environment on Earth in the past. He demonstrated how an electric spark (which is a simulation for lightning) when passed through simple organic gases (similar to the early Earth atmosphere), resulted in the formation of amino acids, which are now known as the building blocks of proteins and the components of living tissues. Theory of Biogenesis The theory of biogenesis presented a strong argument against abiogenesis. This theory states that living things come from living things. Experiments of Francesco Redi and Louis Pasteur disproved the thought of spontaneous generation during their time (figure 5-1). Louis Pasteur argued that life comes from preexisting life and not from nonliving material. However, it should be noted that the "abiogenesis" or "spontaneous generation" that he opposed referred to any modern, existing, fully-formed organisms, not the original generation of life. Louis Pasteur's law of biogenesis contradicted the common belief during his time that cattle dung gives rise to flies, or old clothes with rotten food gives rise to mice. The idea of spontaneous generation was popular until near the seventeenth century. Even famous scientists of that period, such as Descartes, Galileo, and Jan Baptista van Helmont, believed in this theory. CIENCE PIONEER Francesco Redi (1626-1697). Francesco Redi is a physician, a naturalist, and a poet. His works challenged the popular theory of spontaneous generation. He disproved the idea that living things may arise from nonliving things. He also worked on toxicology using viper's venom. He discovered and worked on some parasites that caused disease in humans. SCIENCE CAREER Evolutionary Biologist An evolutionary biologist studies the descent of species and the origin of new species of living things. Working as evolutionary biologist means studying and researching species diversity, their interaction with the environment, their adaptation to change, their ancestors, and their possible origins. This career is important in the field of science because it seeks an answer to the questions about how life began on Earth.Comprehension Test: Superbook – Revelation Theme: Forgiveness, Salvation, and God’s Victory 🔹 Remembering (Items 1–4) What mistake did Chris make at the beginning of the episode? A. He lied to his parents B. He started a fire accidentally C. He broke a valuable item D. He skipped school Answer: B Who are the three main characters in the episode? A. Peter, Mary, and John B. Chris, Joy, and Gizmo C. David, Sarah, and Eli D. Paul, Ruth, and Micah Answer: B Where does Superbook transport the children? A. To the Garden of Eden B. To the time of Jesus’ birth C. To the end times D. To the Exodus Answer: C Who receives visions in heaven during the episode? A. Moses B. Apostle Paul C. Apostle John D. King Solomon Answer: C 🔹 Understanding (Items 5–8) Why does Chris believe his mistake is unforgivable? A. He was punished severely B. He hurt someone intentionally C. He fears the consequences are too great D. He doesn’t understand forgiveness Answer: C What does the descent of the New Jerusalem symbolize? A. The end of all prophecy B. The beginning of war C. Eternal life and God’s promise D. The destruction of the earth Answer: C What lesson does Chris learn from his father? A. That mistakes are permanent B. That forgiveness is earned C. That God’s mercy is unconditional D. That punishment is necessary Answer: C What theme is emphasized throughout the episode? A. Justice and revenge B. Power and control C. Forgiveness and salvation D. Wealth and prosperity Answer: C 🔹 Applying (Items 9–12) If a student feels guilty for a mistake, what lesson from the episode could help them? A. Avoid responsibility B. Seek revenge C. Trust in God’s forgiveness D. Hide the truth Answer: C How might Joy’s experience in heaven help her understand God’s plan? A. She sees the consequences of sin B. She learns about ancient history C. She witnesses God’s ultimate victory D. She meets famous prophets Answer: C What action could reflect the message of salvation in daily life? A. Ignoring others’ mistakes B. Offering forgiveness to someone who hurt you C. Avoiding difficult conversations D. Seeking personal gain Answer: B How could Gizmo’s role support the group’s understanding of Revelation? A. By distracting them with jokes B. By translating the visions C. By guiding them through spiritual truths D. By fixing technical problems Answer: C 🔹 Analyzing (Items 13–16) What contrast is shown between Chris’s fear and the final message of the episode? A. Fear leads to punishment; forgiveness leads to peace B. Fear is stronger than faith C. Mistakes are never forgiven D. God ignores human emotions Answer: A Why is the battle between angels and Satan significant? A. It shows the power of war B. It represents the struggle for heaven C. It symbolizes the victory of good over evil D. It predicts future disasters Answer: C What does the episode suggest about the nature of sin? A. It is always punished B. It can be forgiven through God’s mercy C. It is inherited D. It defines a person’s worth Answer: B How do Chris’s emotions evolve throughout the episode? A. From joy to anger B. From guilt to hope C. From confusion to pride D. From fear to rebellion Answer: B 🔹 Evaluating (Items 17–18) Do you agree that no sin is too great for God’s forgiveness? Why might this be important for viewers? A. Yes, it teaches hope and redemption B. No, some sins are unforgivable C. Yes, but only for believers D. No, forgiveness must be earned Answer: A How effective is the use of apocalyptic imagery in teaching moral lessons? A. It confuses young viewers B. It dramatizes the message of salvation C. It distracts from the core message D. It promotes fear Answer: B 🔹 Creating (Items 19–20) If you were to write a reflection based on this episode, what theme would you focus on? A. The importance of punishment B. The beauty of heaven C. The power of forgiveness D. The fear of judgment Answer: C What title would best capture the message of the episode? A. “The Fire Within” B. “Mercy Wins” C. “The End of Days” D. “Judgment and Justice” Answer: BQ1. A teacher designs a lesson where students compute real-life percentages such as discounts and savings. 👉 A student calculates 15% of 200 to determine savings in a purchase. What is the correct result? A. 20 B. 25 C. 30 D. 35 Q2. In a classroom activity, learners compare numbers to find the highest common factor for grouping materials evenly. 👉 What is the GCF of 24 and 36? A. 6 B. 8 C. 12 D. 18 📘 FRACTIONS, DECIMALS, AND POWERS Q3. A learner converts fractions into percentages for data interpretation. 👉 What is 3/4 expressed as a percentage? A. 50% B. 60% C. 75% D. 80% Q4. A student models exponential growth using repeated multiplication. 👉 What is the value of 252^525? A. 25 B. 30 C. 32 D. 64 📘 ALGEBRA (EQUATIONS AND EXPRESSIONS) Q5. A teacher guides students to solve equations that represent real-life situations. 👉 Solve: 2x+8=202x + 8 = 202x+8=20 A. x = 4 B. x = 6 C. x = 8 D. x = 10 Q6. Students simplify expressions to understand relationships between quantities. 👉 Simplify: 3(x+4)−2x3(x + 4) - 2x3(x+4)−2x A. x + 12 B. x + 4 C. 5x + 4 D. 5x + 12 📘 FUNCTIONS AND GRAPHING Q7. A student analyzes a linear equation to determine its rate of change. 👉 What is the slope of y=3x−5y = 3x - 5y=3x−5? A. -5 B. -3 C. 3 D. 5 Q8. A learner evaluates functions to predict outcomes. 👉 If f(x)=2x+3f(x) = 2x + 3f(x)=2x+3, what is f(4)f(4)f(4)? A. 7 B. 9 C. 11 D. 14 📘 GEOMETRY Q9. Students explore geometric shapes and their properties through visual models. 👉 What is the sum of interior angles of a triangle? A. 90° B. 180° C. 270° D. 360° Q10. A student calculates the area of a classroom table with dimensions 8 cm by 5 cm. 👉 What is the area? A. 26 sq cm B. 30 sq cm C. 40 sq cm D. 48 sq cm 📘 MEASUREMENT AND FIGURES Q11. A learner determines the volume of a cube used in a science experiment. 👉 What is the volume of a cube with side 4 cm? A. 16 cubic cm B. 32 cubic cm C. 48 cubic cm D. 64 cubic cm Q12. Students identify shapes used in design projects. 👉 How many sides does a hexagon have? A. 5 B. 6 C. 7 D. 8 📘 STATISTICS AND PROBABILITY Q13. A teacher helps students interpret data sets using measures of central tendency. 👉 What is the mean of 4, 6, 8, 10, 12? A. 6 B. 8 C. 10 D. 12 Q14. A class experiment involves flipping a fair coin. 👉 What is the probability of getting heads? A. 1/4 B. 1/3 C. 1/2 D. 2/3 📘 WORD PROBLEMS (APPLICATION) Q15. A car travels 180 km in 3 hours during a learning task on speed. 👉 What is its average speed? A. 45 km/h B. 60 km/h C. 75 km/h D. 90 km/h Q16. Students analyze work efficiency in a project. 👉 If 5 workers complete a task in 12 days, how long will 10 workers take? A. 3 days B. 6 days C. 8 days D. 12 days Q17. A student solves a problem involving ratios in a classroom population. 👉 If the ratio of boys to girls is 3:2 and there are 30 students, how many boys are there? A. 12 B. 15 C. 18 D. 20 Q18. A learner determines the duration of a scheduled trip. 👉 A journey starts at 8:30 AM and ends at 11:15 AM. How long is the trip? A. 2 hrs 15 mins B. 2 hrs 30 mins C. 2 hrs 45 mins D. 3 hrs 15 mins Q19. A student computes simple interest for financial literacy. 👉 What is the simple interest on ₱1000 at 5% for 2 years? A. ₱50 B. ₱75 C. ₱100 D. ₱150 Q20. A learner solves a perimeter problem involving a rectangle. 👉 A rectangle has a length of 12 cm and perimeter of 34 cm. What is the width? A. 5 cm B. 7 cm C. 10 cm D. 11 cm ✅ ANSWER KEY (BASED ON YOUR REVIEWER) (All verified from your uploaded file) [ilide.info...002acd4e5a | PDF] QAnswer1C2C3C4C5B6A7C8C9B10C11D12B13B14C15B16B17C18C19C20A