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During Travels FIB 1
Quiz by Lia Poole
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During Your Travels2 ovaries → follicles → oocyte Follicles: made of 1 oocyte + theca cells and granulosa cells Theca cells - make testosterone Granulosa cells - make estradiol Oocytes: fixed amount per lifetime (~400) Ovulation - when oocyte leaves the follicle Fertilization: Connects ovaries to uterus Open-system: fallopian tubes are open to peritoneal space → easy to get infection! NOTE: Endometriosis: when endometrial cells are found in places other than uterus Retrograde menstruation: shedding of endometrium → goes up to fallopian tube (instead of down to cervix → vagina) → enters peritoneal space → travels Endometrium: hormone-sensitive mucosal lining of the endometrium Non-pregnant menstrual cycle: bleeding = shedding of endometrium Pregnancy: endometrium doesn’t shed → implantation occurs here Myometrium: smooth muscle layer of uterus → contracts during period to shed endometrium → cramps Continuous with cervix Dilates during menses and during birth What is a Hurricane, Typhoon, or Tropical Cyclone? The terms "hurricane" and "typhoon" are regionally specific names for a strong "tropical cyclone". A tropical cyclone is the generic term for a non-frontal synoptic scale low-pressure system over tropical or sub-tropical waters with organized convection (i.e. thunderstorm activity) and definite cyclonic surface wind circulation (Holland 1993). Tropical cyclones with maximum sustained surface winds of less than 17 m/s (34 kt, 39 mph) are usually called "tropical depressions" (This is not to be confused with the condition mid-latitude people get during a long, cold and grey winter wishing they could be closer to the equator). Once the tropical cyclone reaches winds of at least 17 m/s (34 kt, 39 mph) they are typically called a "tropical storm" or in Australia a Category 1 cyclone and are assigned a name. If winds reach 33 m/s (64 kt, 74 mph), then they are called: "hurricane" (the North Atlantic Ocean, the Northeast Pacific Ocean east of the dateline, or the South Pacific Ocean east of 160E) "typhoon" (the Northwest Pacific Ocean west of the dateline) "severe tropical cyclone" or "Category 3 cyclone" and above (the Southwest Pacific Ocean west of 160°E or Southeast Indian Ocean east of 90°E) "very severe cyclonic storm" (the North Indian Ocean) "tropical cyclone" (the Southwest Indian Ocean) Coriolis Effect The Coriolis Effect—the deflection of an object moving on or near the surface caused by the planet’s spin—is important to fields, such as meteorology and oceanography. Storm Approaching Southeast Asia Because of the Coriolis Effect, hurricanes spin counterclockwise in the Northern Hemisphere, while these types of storms spin clockwise in the Southern Hemisphere. This Northern Hemisphere storm, approaching Southeast Asia, is spinning counterclockwise. Earth is a spinning planet, and its rotation affects climate, weather, and the ocean through the Coriolis Effect. Named after the French mathematician Gaspard Gustave de Coriolis (born in 1792), the Coriolis Effect refers to the curved path that objects moving on Earth’s surface appear to follow because of the spinning of the planet. As Earth turns, points near the equator—countries like Ecuador and Kenya—are moving much faster than places near the planet’s poles. This is because Earth is shaped like a marble: Its circumference is larger near its middle (the equator) than near its top and bottom. All places on Earth experience a day that is about 24 hours long, but points near the equator have to travel longer distances in the same period of time, which means that those places move faster. Scientists say these points have more “angular momentum.” This is why rockets are usually launched from places near the equator, like Cape Canaveral, Florida, United States. Such locations give rockets a large initial speed, which helps them get into orbit using the least possible amount of fuel. The Coriolis Effect influences wind patterns, which in turn dictate how ocean currents move. Imagine wind near the equator flowing to the north. That wind starts with a certain speed due to Earth’s rotation (near the equator, Earth rotates at a speed of roughly 1,600 kilometers per hour (1,000 miles per hour) from west to east). As the wind travels north toward the North Pole, it moves over parts of Earth that are rotating progressively more slowly. Since the wind retains its angular momentum, it keeps moving from west to east, overtaking the part of Earth turning more slowly below it. As a result, the wind appears to bend to the east (that is, to the right). This is the Coriolis Effect in action. Wind flowing south from the equator would likewise bend to the east. This effect is responsible for many meteorological and oceanographic phenomena. For instance, due to the Coriolis Effect, hurricanes in the Northern Hemisphere spin in a counterclockwise direction, while hurricanes in the Southern Hemisphere (known as cyclones) spin in a clockwise direction. Ocean-circling currents known as “gyres” also spin in spiral patterns thanks to the Coriolis Effect. There is an urban legend that water in toilets spins in opposite directions in the Northern and Southern Hemispheres because of the Coriolis Effect. But that isn't true—a toilet bowl is too small for the effect to be observed. Instead, other factors like the shape of the toilet bowl and the direction that the water enters are largely responsible for how the flushing water moves.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.Q1. 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] QAnswer1C2C3C4C5B6A7C8C9B10C11D12B13B14C15B16B17C18C19C20ADuring Travel Free Text QuizAre you feeling adventurous? Then spread your wings and travel km southeast of Taiwan, where you'll discover a part of the country unlike any other. Known as Lanyu* or Orchid Island, this tiny drop of earth in the Pacific* is home to the Tao*, Taiwan's only ocean-dependent indigenous* tribe. The Tao people are the people of the "alibangbang*," or flying fish, one of Mother Nature's greatest gifts. The annual flying fish season is at the heart of Tao culture and comes with many traditions and taboos. Spanning around eight months of the year, the flying fish season involves multiple customs and rituals. A special religious ceremony, usually held in February, marks the beginning of this important period and ensures a prosperous and plentiful season. Tao boat captains head to the beach with elaborate costumes on. There, they pray and make sacrifices to appeal to the flying fish spirits to bless their tribe, and to the gods for good fortune and courage on the dangerous seas. Although local fishers can catch flying fish between March and June, they are restricted to capturing only those found in shallow waters. During this period, netting* any deep-sea or coral reef fish is not permitted. If the fishers didn't follow this rule, various species would not have the time needed to recover their numbers, and harmony with nature would be lost. At the end of June, the annual sea harvest concludes with another ceremony to express gratitude and respect for nature's endless* cycles. After the ceremony, flying fish can no longer be caught. Therefore, from July onward*, only dried and stored alibangbang can be eaten. Furthermore, after the Mid-Autumn Festival*, even these dried fish must no longer be consumed. This custom seems to have been engraved in the locals’ hearts* and is believed to prevent ill luck and guarantee abundance in seasons to come. Just as the locals are concluding their scaly* harvest for the year, scores of tourists begin arriving on Lanyu for the summer season. To make sure you don't feel like a fish out of water, here are some basic guidelines to help you enjoy your stay while respecting local traditions. First, it is considered a taboo to touch or photograph the Tao fishing boats without permission since it would bring bad luck. Also, you should never enter locals’ private gardens uninvited* to view or sample drying fish. Additionally, avoid asking about future fishing trips, as locals believe that the gods may be angered and that the harvest may be spoiled as a consequence. Finally, as a sign of respect for the local community and environment, don't snorkel* or dive in the island's traditional fishing waters during the flying fish season. Such activities can disturb the local ecosystem. The customs and taboos mentioned above reflect the importance of preserving and cherishing natural resources and local culture. The ancient wisdom of Lanyu’s Tao inhabitants thus appears to give us much to learn about living in harmony with this planet. Everyone is, after all, in the same boat under nature's wing.The film begins as a journey to film the largest animal on the planet, the blue whale. But during the journey the filmmakers (journalist Craig Leeson and environmental activist Tanya Streeter) make the shocking discovery of a huge, thick layer of plastic floating in the middle of the Indian Ocean. This prompts them to travel around the world to look at other areas that have been affected. In total, they visited 20 locations around the world during the four years it took them to make the film. The documentary premiered in 2016, and is now on streaming services such as Netflix.It’s very clear that a lot of research went into the film. There are beautiful shots of the seas and marine life. These are contrasted with scenes of polluted cities and dumps full of plastic rubbish. We see how marine species are being killed by all the plastic we are dumping in the ocean. The message about our use of plastic is painfully obvious. But the film doesn’t only present the negative side. In the second half, the filmmakers look at what we can do to reverse the tide of plastic flowing around the world. They present short-term and long-term solutions. These include avoiding plastic containers and ‘single-use’ plastic products as much as possible. Reuse your plastic bags and recycle as much as you can. The filmmakers also stress the need for governments to work more on recycling programmers, and look at how technology is developing that can convert plastic into fuel We make a staggering amount of plastic. In terms of plastic bags alone, we use five hundred billion worldwide annually. Over 300 million tons of plastic are produced every year, and at least 8 million of those are dumped into the oceans. The results are disastrous, but it isn’t too late to change. Once you’ve seen A Plastic Ocean, you’ll realize the time is now and we all have a role to play.