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The Dot
Quiz by Maria Leon
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Title: The Adventures of Max and the Magical Computer (Shortened Version) Once upon a time, in a small town called Techville, there lived a curious fourth-grader named Max. One day, while exploring his grandma’s attic, he found an old, dusty computer. As he cleaned it, the screen lit up, and a cheerful voice said, “Hello, Max! I’m Compy, your magical computer. Let’s learn about operating systems, files, and folders!” Max was thrilled. “A talking computer? Let’s go!” Chapter 1: The World of Windows Compy explained, “I run on Windows, the brain of the computer. It uses a Graphical User Interface (GUI), so you can interact with me using icons, menus, and buttons. Let’s start by changing my desktop background—the image on the screen.” Max chose a spaceship picture. “Cool! Can I add a screen saver too?” “Of course!” said Compy. “It’s an image that appears when I’m inactive. Try this swirling galaxy!” Max set the screen saver and giggled as it appeared. “This is fun!” Chapter 2: Organizing with Folders and Files Compy’s screen filled with random icons. “Oh no! My files are a mess. Can you help?” “Sure! What are files and folders?” asked Max. “A file is information, like a picture or document. Files have names, like ‘Homework.docx.’ The part after the dot, like .docx, is the file extension. It tells you the file type,” Compy explained. Max pointed to “Game.exe.” “So, this is a program file?” “Yes!” said Compy. “To organize, we use folders—like drawers for files. You can even put folders inside folders!” Max created a “School” folder, added his homework files, and made a “Projects” folder inside it. “Now everything’s neat!” Chapter 3: The File Explorer Adventure Compy’s screen flickered. “Some files are missing. Let’s use Windows Explorer to find them. It’s like a map for files and folders.” Max opened Windows Explorer and saw a tree-like list of folders. “This is like a tree with branches!” “Exactly!” said Compy. “Search for the missing files and move them to the right folders.” Max found the files in “Downloads” and moved them. “I feel like a computer detective!” Chapter 4: The Final Challenge Compy’s screen turned into a game board. “Time for a quiz! What’s the purpose of an operating system? How do you change the desktop background? What’s the difference between a file and a folder?” Max answered all the questions correctly, and fireworks lit up the screen. “Congratulations, Max! You’re a computer whiz!” The End Max smiled. “Thanks, Compy! I can’t wait to teach my friends!” Compy replied, “Remember, Max, learning is an adventure. Keep exploring!” As Max turned off the computer, he knew his journey into technology had just begun.
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.
Am I all right? Why do doctors wear green or blue clothes while doing an operation? Look at the first picture. Stare at the dot for 10 seconds. Then look at the second picture. What can you see? What you see is called afterimage. Afterimage is an image that continues to appear in people's vision after the image has disappeared. The blood is red. But if you look at blood for a long time, you will be affected by afterimage. Doctors change their scrubs from white into green because green may help doctors see the room clearly after staring at blood for a long time.
Volcanic Eruptions Earthquakes are sometimes used in predicting volcanic eruptions. An earthquake alerts the authorities to observe a nearby volcano. When a volcano shows signs of eruption, evacuation of the nearby villages should be immediately implemented Active volcanoes are the ones that are usually closely monitored by volcanologists These are commonly located along plate boundaries specifically along the Pacific Ring of Fire. The Philippines is one of the countries that have a lot of volcanoes. The Pacific Ring of Fire is a region in the Pacific where most active volcanoes are located. The dots in the figure represent the active volcanoes in the world. Active volcanoes are those that have a recorded volcanic eruption in their history. There are two types of volcanic eruptions depending on the magma composition: the explosive type and the Hawaiian or nonexplosive type. The types of eruptions depend on the viscosity and amount of dissolved gases in a magma. Explosive eruptions have magma that is highly viscous and contains large amounts of dissolved gases. On the other hand, the Hawaiian type or nonexplosive eruptions have magma that has low viscosity and low amount of dissolved gases. Viscosity determines the ability of magma to flow. The lower the viscosity, the easier it is for the magma to flow. This viscosity of the magma is affected by silica content and temperature. Low-temperature magma with high silica content are highly viscous, high-temperature magma with low silica content has low viscosity. Therefore, in explosive eruptions, the magma could not easily get out of the vent. This impedes the remaining magma below, causing an increase of pressure. Once the pressure is released. an explosive eruption occurs. Volcanic Hazards Explosive volcanic eruptions release pyroclastic flows and materials that are considered to pose the greatest threat. The pyroclastic flows often render people in a shock state since they become overwhelmed with what they are seeing Too much pyroclastic material (eg., tephra), may bury people and severely damage the agricultural land and livestock. Damaged agricultural land could lead to famine. After an eruption, pyroclastic materials that have settled along the slope of the volcano get mixed with rainwater. This would start a mudflow or lahar moving at high speed, destroying everything along its path. Poisonous gases are also released by volcanoes. These gases can be lethal to people, animals, and plants. Just like earthquakes, violent volcanic eruptions may also happen underwater, which can trigger a tsunami.
Many of water’s biological functions stem from its chemical struc- ture. Recall that in the water molecule, H2O, the hydrogen and oxygen atoms share electrons to form covalent bonds. However, these atoms do not share the electrons equally. The oxygen atom has a greater ability to attract electrons to it because it pulls hydrogen’s electrons towards its nucleus. As a result, as shown in Figure 2-8, the region of the molecule where the oxygen atom is located has a partial negative charge, denoted with a , while the regions of the molecule where each of the two hydrogen atoms are located have partial positive charges, each of which are denoted with a . Thus, even though the total charge on a water molecule is neutral, the charge is unevenly distributed across the water molecule. Because of this uneven distribution of charge, water is called a polar compound. Notice also in Figure 2-8 that the three atoms in a water mole- cule are not arranged in a straight line as you might expect. Rather, the two hydrogen atoms bond with the single oxygen atom at an angle. SECTION 3 OBJECTIVES ● Describe the structure of a water molecule. ● Explain how water’s polar nature affects its ability to dissolve substances. ● Outline the relationship between hydrogen bonding and the different properties of water. ● Identify the roles of solutes and solvents in solutions. ● Differentiate between acids and bases. VOCABULARY polar hydrogen bond cohesion adhesion capillarity solution solute solvent concentration saturated solution aqueous solution hydroxide ion hydronium ion acid base pH scale buffer Copyright © by Holt, Rinehart and Winston. All rights reserved. (a) Electron cloud model (b) Space-filling model H H O The oxygen region of the water molecule is weakly negative, and the hydrogen regions are weakly positive. Notice the different ways to represent water, H2O. You are familiar with the electron cloud model (a). The space- filling model (b) shows the three- dimensional structure of a molecule. FIGURE 2-8 40 CHAPTER 2 Hydrogen bond H H H H H H H H H O O O O O O H H H H H – – – – – – – + + + + + + + + + + + + + + The dotted lines in this figure represent hydrogen bonds. A hydrogen bond is a force of attraction between a hydrogen atom in one molecule and a negatively charged region or atom in a second molecule. FIGURE 2-10 The positive region of a water molecule attracts the negative region of an ionic compound, such as the Cl portion of NaCl. Similarly, the negative region of the water molecule attracts the positive region of the compound—the Na portion of NaCl. As a result, NaCl breaks apart, or dissolves, in water. FIGURE 2-9 CI– Na+ H2O + + – – Solubility of Water The polar nature of water allows it to dissolve polar substances, such as sugars, ionic compounds, and some proteins. Water does not dissolve nonpolar substances, such as oil because a weaker attraction exists between polar and nonpolar molecules than between two polar molecules. Figure 2-9 shows how water dissolves the ionic compound sodium chloride, NaCl. In your body, ions, such as sodium and chloride, are essential to bodily func- tions, such as muscle contraction and transmission of impulses in the nervous system. In fact, dissolved, or dissociated ions, are pre- sent in all of the aqueous solutions found in living things and are important in maintaining normal body functions. HYDROGEN BONDING The polar nature of water also causes water molecules to be attracted to one another. As is shown in Figure 2-10, the positively charged region of one water molecule is attracted to the negatively charged region of another water molecule. This attraction is called a hydrogen bond. A hydrogen bond is the force of attraction between a hydrogen molecule with a partial positive charge and another atom or molecule with a partial or full negative charge. Hydrogen bonds in water exert an attractive force strong enough so that water “clings” to itself and some other substances. Hydrogen bonds form, break, and reform with great frequency. However, at any one time, a great number of water molecules are bonded together. The number of hydrogen bonds that exist depends on the state that water is in. If water is in its solid state all its water molecules are hydrogen bonded and do not break. As water liquifies, more hydrogen bonds are broken than are formed, until an equal number of bonds are formed and broken. Hydrogen bonding accounts for the unique properties of water, some of which we will examine further. These properties include cohesion and adhesion, the ability of water to absorb a relatively large amount of energy as heat, the ability of water to cool surfaces through evaporation, the density of ice, and the ability of water to dissolve many substances.
Lesson 2: Plate Tectonics There are a few handfuls of major plates and dozens of smaller, or minor, plates. Six of the majors are named for the continents embedded within them, such as the North American, African, and Antarctic plates. Though smaller in size, the minors are no less important when it comes to shaping the Earth. The tiny Juan de Fuca plate is largely responsible for the volcanoes that dot the Pacific Northwest of the United States. The plates make up Earth's outer shell, called the lithosphere. (This includes the crust and uppermost part of the mantle.) Churning currents in the molten rocks below propel them along like a jumble of conveyor belts in disrepair. Most geologic activity stems from the interplay where the plates meet or divide. The movement of the plates creates three types of tectonic boundaries: convergent, where plates move into one another; divergent, where plates move apart; and transform, where plates move sideways in relation to each other. They move at a rate of one to two inches (three to five centimeters) per year. Convergent BoundariesWhere plates serving landmasses collide, the crust crumples and buckles into mountain ranges. India and Asia crashed about 55 million years ago, slowly giving rise to the Himalaya, the highest mountain system on Earth. As the mash-up continues, the mountains get higher. Mount Everest, the highest point on Earth, may be a tiny bit taller tomorrow than it is today. These convergent boundaries also occur where a plate of ocean dives, in a process called subduction, under a landmass. As the overlying plate lifts up, it also forms mountain ranges. In addition, the diving plate melts and is often spewed out in volcanic eruptions such as those that formed some of the mountains in the Andes of South America. At ocean-ocean convergences, one plate usually dives beneath the other, forming deep trenches like the Mariana Trench in the North Pacific Ocean, the deepest point on Earth. These types of collisions can also lead to underwater volcanoes that eventually build up into island arcs like Japan. Divergent Boundaries At divergent boundaries in the oceans, magma from deep in the Earth's mantle rises toward the surface and pushes apart two or more plates. Mountains and volcanoes rise along the seam. The process renews the ocean floor and widens the giant basins. A single mid-ocean ridge system connects the world's oceans, making the ridge the longest mountain range in the world. On land, giant troughs such as the Great Rift Valley in Africa form where plates are tugged apart. If the plates there continue to diverge, millions of years from now eastern Africa will split from the continent to form a new landmass. A mid-ocean ridge would then mark the boundary between the plates. Transform Boundaries The San Andreas Fault in California is an example of a transform boundary, where two plates grind past each other along what are called strike-slip faults. These boundaries don't produce spectacular features like mountains or oceans, but the halting motion often triggers large earthquakes, such as the 1906 one that devastated San Francisco.
Create a quiz over the following topic: Visual Art: Elements of Art, Music, Dance, and Drama A. Elements of Visual Art: There are seven primary elements of art (line, shape, form, space, color, value, texture), the components from which any artwork can be built. Any artwork that can stand the test of time will contain these elements. Element 1 – Line Line is the most fundamental concept in art. This element leads the viewer’s eye around the artwork, communicating the piece’s meaning to the viewer. Without this element, none of the other elements could exist. At its most basic foundation, a line is simply a moving dot. When these dots overlap, the line is solid. When they do not overlap, the line is dotted or dashed. These “incomplete” lines can represent movement or hidden material behind the primary figure. Repeated lines create patterns, and these patterned lines create a rhythm for the work. Lines can be: • Curved • Zigzag • Horizontal • Vertical • Diagonal • Implied • Gesture • Varied thicknesses • Varied lengths Examples of lines and their meaning in art: Thick Lines – give emphasis and advance Thin Lines – recede and cause the eye to “lose interest” Straight Lines – rarely found in nature, thus are inorganic. Straight lines are dynamic and mechanistic. Curved Lines – gently change directions and have no sharp angles, suggesting calmness – Curved lines are more readily found in nature and are thus considered organic lines Works with pronounced curved lines include Van Gogh’s The Starry Night (1889). Note the movement across the sky.