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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 â˘(-9.8m/s2) â˘d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) â˘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.
Are 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. Â Â Â Â Â Â Â 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. Â Â Â Â Â Â Â 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. Â Â Â Â Â Â Â 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 localsâ hearts* and is believed to prevent ill luck and guarantee abundance in seasons to come. Â Â Â Â Â Â Â 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 localsâ 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. Â Â Â Â Â Â Â The customs and taboos mentioned above reflect the importance of preserving and cherishing natural resources and local culture. The ancient wisdom of Lanyuâs 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.
Generate all of these 25 questions Part A: Each correct answer is worth 5. 1. The regular pentagon shown has a side length of 2 cm. The perimeter of the pentagon is (A) 2 cm (B) 4 cm (C) 6 cm (D) 8 cm (E) 10 cm 2 cm 2. The faces of a cube are labelled with 1, 2, 3, 4, 5, and 6 dots. Three of the faces are shown. What is the total number of dots on the other three faces? (A) 6 (B) 8 (C) 10 (D) 12 (E) 15 3. The equation that best represents \a number increased by _ve equals 15" is (A) n ô 5 = 15 (B) n _ 5 = 15 (C) n + 5 = 15 (D) n + 15 = 5 (E) n _ 5 = 15 4. The line graph shows the number of bobbleheads sold at a store each year. The sale of bobbleheads increased the most between (A) 2016 and 2017 (B) 2017 and 2018 (C) 2018 and 2019 (D) 2019 and 2020 (E) 2020 and 2021 Number of 2016 2017 2018 2019 2020 Year Sale of Bobbleheads 2021 Bobbleheads 20 40 60 80 5. Starting at 72, Aryana counts down by 11s: 72; 61; 50; : : : . What is the last number greater than 0 that Aryana will count? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 6. In the diagram, \ABC = 90_. The value of x is (A) 68 (B) 23 (C) 56 (D) 28 (E) 26 Day of the Week 44° x° A B C x° 7. Which of the following values is closest to zero? (A) ô1 (B) 5 4 (C) 12 (D) ô4 5 (E) 0:9 Grade 8 8. A jar contains 267 quarters. One quarter is worth $0.25. How many quarters must be added to the jar so that the total value of the quarters is $100.00? (A) 33 (B) 53 (C) 103 (D) 133 (E) 153 9. A package of 8 greeting cards comes with 10 envelopes. Kirra has 7 cards but no envelopes. What is the smallest number of packages that Kirra needs to buy to have more envelopes than cards? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 10. For the points in the diagram, which statement is true? (A) e > c (B) b < d (C) f > b (D) a < e (E) a > c y x (e, f ) (a, b) (c, d ) Part B: Each correct answer is worth 6. 11. The 26 letters of the English alphabet are listed in an in_nite, repeating loop: ABCDEFGHIJKLMNOPQRSTUVWXY ZABC : : : What is the 258th letter in this sequence? (A) V (B) W (C) X (D) Y (E) Z 12. A public holiday is always celebrated on the third Wednesday of a certain month. In that month, the holiday cannot occur on which of the following days? (A) 16th (B) 22nd (C) 18th (D) 19th (E) 21st 13. A circular spinner is divided into three sections. An arrow is attached to the centre of the spinner. The arrow is spun once. The probability that the arrow stops on the largest section is 50%. The probability it stops on the next largest section is 1 in 3. The probability it stops on the smallest section is (A) 1 4 (B) 2 5 (C) 1 6 (D) 2 7 (E) 3 10 14. A positive number is divisible by both 3 and 4. The tens digit is greater than the ones digit. How many positive two-digit numbers have this property? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 15. A rectangular pool measures 20 m by 8 m. There is a 1 m wide walkway around the outside of the pool, as shown by the shaded region. The area of the walkway is (A) 56 m2 (B) 60 m2 (C) 29 m2 (D) 52 m2 (E) 50 m2 20 m 8 m 1 m Grade 8 16. The results of asking 50 students if they participate in music or sports are shown in the Venn diagram. What percentage of the 50 students do not participate in music and do not participate in sports? (A) 0% (B) 80% (C) 20% (D) 70% (E) 40% Music Sports 15 5 20 17. There are 2 3 as many golf balls in Bin F as in Bin G. If there are a total of 150 golf balls, how many fewer golf balls are in Bin F than in Bin G? (A) 15 (B) 30 (C) 50 (D) 60 (E) 90 18. In the sequence shown, Figure 1 is formed using 7 squares. Each _gure after Figure 1 has 5 more squares than the previous _gure. What _gure has 2022 squares? (A) Figure 400 (B) Figure 402 (C) Figure 404 (D) Figure 406 (E) Figure 408 Figure 1 Figure 2 Figure 3 19. Mateo's 300 km trip from Edmonton to Calgary passed through Red Deer. Mateo started in Edmonton at 7 a.m. and drove until stopping for a 40 minute break in Red Deer. Mateo arrived in Calgary at 11 a.m. Not including the break, what was his average speed for the trip? (A) 83 km/h (B) 94 km/h (C) 90 km/h (D) 95 km/h (E) 64 km/h 20. Equilateral triangle ABC has sides of length 4. The midpoint of BC is D, and the midpoint of AD is E. The value of EC2 is (A) 7 (B) 6 (C) 6:25 (D) 8 (E) 10 Part C: Each correct answer is worth 8. 21. The positive factors of 6 are 1, 2, 3, and 6. There are two perfect squares less than 100 that have exactly _ve positive factors. What is the sum of these two perfect squares? (A) 177 (B) 80 (C) 145 (D) 52 (E) 97 22. In the list p; q; r; s; t; u; v, each letter represents a positive integer. The sum of the values of each group of three consecutive letters in the list is 35. If q + u = 15, then p + q + r + s + t + u + v is (A) 85 (B) 70 (C) 80 (D) 90 (E) 75 Grade 8 23. The net shown is folded to form a cube. An ant walks from face to face on the cube, visiting each face exactly once. For example, ABCFED and ABCEFD are two possible orders of faces the ant visits. If the ant starts at A, how many possible orders are there? (A) 24 (B) 48 (C) 32 (D) 30 (E) 40 A D B C E F 24. The number 385 is an example of a three-digit number for which one of the digits is the sum of the other two digits. How many numbers between 100 and 999 have this property? (A) 144 (B) 126 (C) 108 (D) 234 (E) 64 25. Student A, Student B, and Student C have been hired to help scientists develop a new avour of juice. There are 4200 samples to test. Each sample either contains blueberry or does not. Each student is asked to taste each sample and report whether or not they think it contains blueberry. Student A reports correctly on exactly 90% of the samples containing blueberry and reports correctly on exactly 88% of the samples that do not contain blueberry. The results for all three students are shown below. Student A Student B Student C Percentage correct on samples 90% 98% (2m)% containing blueberry Percentage correct on samples 88% 86% (4m)% not containing blueberry Student B reports 315 more samples as containing blueberry than Student A. For some positive integers m, the total number of samples that the three students report as containing blueberry is equal to a multiple of 5 between 8000 and 9000. The sum of all such values of m is (A) 45 (B) 36 (C) 24 (D) 27 (E) 29
Broken windows are covered. Floorboards are patched and doors screwed back on. The road that was ruined by German tanks is shovelled and raked smooth. Boot-shaped bruises turn yellow then fade and disappear. Flowers grow and spread across the ugly German footprints stomped into garden beds. The village looks pretty once more. School stops for the summer and everyone is put to work on the kolkhoz, the village farm. Women and big boys begin harvesting the barley crops in the outer fields. The biggest girls milk the cows, morning and night, and keep the barns clean. Old Nikolay mends ploughs, horse harnesses, pitchforks and scythes in his workshop. Anna Pushinka teaches Yelena and her friends how to get the honey from the beehives that are scattered through the orchards. I am in charge of collecting eggs. My friends Olga and Nina help. Olga and Nina are five, a year younger than me. They are twins and look exactly alike, except Ninaâs nose is a little bit crooked from when she fell out of bed and squashed it sideways on the floor. The hens, ducks and geese wander free in the summer, so collecting eggs is like a treasure hunt and takes hours. Catching the hens for their daily hugs takes even longer, but I think itâs important because hugs make everyone happy and happy hens lay bigger eggs. Olga says Iâm the best hen-hugger in all of Russia. Nina says Iâll be the best cow-hugger, too, when my arms grow longer. But good hugs have nothing to do with the size of your arms. Itâs all to do with the size of your heart. When we are done with the hens, Olga, Nina and I can spend the rest of the day doing whatever we like. We climb the apricot trees, chase squirrels, lie in the meadow marvelling at how hot Ushankaâs black fur becomes in the sunshine, make daisy chains and race little boats of bark in the stream. I teach Olga and Nina the alphabet and we use charcoal to write our letters and our names all over the village â on doors and walls and the freshly cut ends of firewood. In between, I practise my knots. In case the German princemonsters return. I slip into Old Nikolayâs workshop and tie knots in the harnesses hanging on the walls. I wander into gardens where the washing is hung out to dry and tie knots in the laces on pants and smocks. I creep up behind Anna Pushinka and tie knots in her apron strings. I find baling twine in the hay shed and tie my own ankles together. I do such a good job of these last knots that I canât get them undone. I have to jump all the way to Olga and Ninaâs house and ask them to cut me free with their mamaâs knife. At the end of each day, Ushanka and I run out into the distant barley fields to meet Mama. This is my favourite part of the day, because Mama always shouts, âLittle Rabbit!â and smothers my head with kisses. And as we walk home, we sing. Everyone â women, big boys and me. I love to sing. Almost as much as I love to be kissed by Mama. Sometimes one of the boys, Mikhail, has his balalaika with him. He takes the instrument out from beneath the sheaves of barley piled high on the wagon and plays music. We sing about forests and orchards and people who find their true love. As we walk home, arm in arm, my heart fills with happiness and my belly swells with pride that I am allowed to sing along with the big boys. And I can almost forget about the German prince-monsters and their lies about Russia and their big ugly boots. Almost. But today, when Mikhail reaches for his balalaika, I see other things hiding beneath the barley sheaves. Three of the mamas rush forward and cover them up, but itâs too late. I know they are there. Iâve already seen them. Rifles. Lots of rifles. Mikhail hugs his balalaika to his chest and blushes. âSo play!â cries Mama, her voice oddly loud and high. âLetâs play Sashaâs favourite song, âThe Little Birch Treeâ.â So Mikhail plays and everyone sings about the lovely birch tree with its curly leaves and the branches that will be turned into silver flutes. They sing too quickly, too loudly, and as they sing and walk, they cast nervous sideways glances at me. âItâs alright,â I say, when the song comes to an end. âI didnât see the rifles.â Mama nods and smiles, and I know it was the right thing to say. But I did see the rifles. And I think about Yelena wanting to get lots of guns and dynamite for the Partisans so they can shoot the Germans and blow them into thousands of tiny pieces, and Mama looking as though she agreed, and I know this is what the mamas and the big boys are doing. As well as harvesting, they are helping the Partisans. Three days later, I wake before dawn and I am all alone. Yelena is always here beside me when I wake. But not this morning. I climb down from our bed above the stove. Mama is filling a cloth sack with bread. She ties it closed with a piece of string and hands it to Yelena. âStay out of sight,â says Mama. âAnd donât return until after dark.â âWhereâs she going?â I ask. âNowhere,â snaps Mama. âThen why does she need all that bread?â I ask. âThereâs nothing left for us.â Mama baked four loaves last night and she has stuffed them all into the sack. Yelena opens her mouth, but before she can speak, Mama shoves her out the door and sends her on the way to nowhere. Mama turns and stares at me, her blue, blue cornflower eyes wide with worry. âI know,â I say, flopping down on the bench. âI didnât see any bread.â Mama sits beside me and takes my hand. âAnd . . .?â she prods, obviously waiting for more. I puzzle for a while, then say, âAnd I donât have a sister called Yelena.â Mama laughs, softly and with a little bit of sadness around the edges. âSweet Little Rabbit! You do have a sister called Yelena.â âI do?â I ask, now confused. âI havenât seen the rifles or the bread, but I have seen Yelena?â âYes.â Mama smiles and the magic makes me smile, too. And I am glad that Yelena is real because I love her very much. âYelena is real,â Mama explains, âbut she does not carry sacks of bread into the forest for the Partisans.â âOf course not!â I shout, slapping my forehead. âBecause there is no bread!â Mama laughs loudly now, with not a hint of sadness. She hugs me, pressing me against her warm, loving heart, covering my head with kisses. âClever Little Rabbit,â she murmurs, and then, in barely a whisper, âYour papa would be so proud.â When I wake the next morning, Yelena is sleeping beside me, her mouth open, her braided hair unravelling. Mama is serving kasha to a strange woman seated at our table. I crawl down from above the stove and slide along the bench beside her. I stare at her pants, her tunic, the rope she is using as a belt and her big boots. Sheâs dressed like a man! And thereâs a rifle leaning against the wall near the door. âHello,â I say. âIâm Sasha.â The woman doesnât reply. She just shovels down her kasha. I line my four wooden bears along the table in front of her bowl and say, âThese are my bears: Big Bear, Medium Bear, Little Bear and Even Littler Bear.â âHello, Sasha. Hello, bears.â She smiles but she doesnât tell me her name. âWhy are you dressed like a man?â I ask, tugging at the sleeve of her tunic. âBecause menâs clothes make it easier to run and climb and crawl and shoot,â she says. âYouâre a Partisan!â I gasp. âBut sheâs not real,â says Mama, placing a bowl of kasha before me. âIs the kasha real?â I ask. Mama laughs. âYes, Little Rabbit.â Iâm glad the food is real, because Iâm hungry. But Iâm disappointed that the woman is not real. I was going to ask if I could use her rope-belt to tie her ankles together. For practice. But if sheâs not real, then the rope and her ankles arenât either. The woman finishes her kasha, hangs her rifle over her shoulder, kisses Mama on the cheek then slips out the door. I run to the window to watch her leave, but by the time I get there, sheâs gone. Vanished. âBecause sheâs not real,â I whisper. A week later, Mama and I are working in the garden. We sing as we weed between the flowers and pluck caterpillars from the vegetables. Anna Pushinka is picking strawberries in her garden and wanders over. âTaste these,â she says, holding out the basket. Mama reaches in and takes out a fat strawberry and a tiny piece of folded paper. The strawberry goes into her mouth, the paper into her pocket. âWhatâs on the paper?â I ask. âPaper?â Anna Pushinka replies with a wave of her hand. âGoodness, Sasha! Who has money for paper? These are lean times. We must choose between paper for writing and noodles for our soup. And I always choose noodles.â She chuckles and I know the paper is yet another thing that is not real. That night, Mama slips the paper to Yelena, but she drops it on the floor. I pick it up for her, and I see that there are tiny words and numbers written all over it. I wish I could read better. Iâm desperate to know what it says. Or rather, what it doesnât say, because itâs not real. Later, when Mama has tucked us into our bed above the stove and Ushanka has wrapped herself around the top of my head, I ask Yelena, âWhatâs on the paper?â âWhat paper?â says Yelena. âThe paper that isnât real,â I reply. Yelena stares at me, nibbling her lip, then whispers, âA message for the Partisans. Stuff about where the Germans have their headquarters and when their trains are travelling and where they store their ammunition.â âWhy?â âSo the Partisans can blow them up.â Yelena grabs my arm. âBut donât tell anyone. Itâs a secret.â âWhatâs a secret?â I ask. âThe message.â âWhat message?â I say, my eyes wide. Yelena laughs. âGood boy, Sasha.â My belly swells with pride. I know how to play this game. âHow are your knots coming along?â asks Yelena. âGood! Yesterday, I crept into the dairy and tied knots in the apron strings of all the girls who were milking and only one of them noticed. Today, I tied Olgaâs ankles together with Mamaâs embroidery thread and just now, while you were taking a bath, I tied the sleeves of your blouse together in an enormous knot.â Yelena rolls her eyes, then says, âIâll see if I can find you some rope for practising.â âPractising what?â I ask. âYour knots,â she says. âWhat knots?â Yelena, my big sister who is twelve and always serious t
Number that comes after Quiz
CARBOHYDRATES Carbohydrates are organic compounds composed of carbon, hydrogen, and oxygen in a ratio of about one carbon atom to two hydrogen atoms to one oxygen atom. The number of carbon atoms in a carbohydrate varies. Some carbohydrates serve as a source of energy. Other carbohydrates are used as structural materials. Carbohydrates can exist as monosaccharides, disaccharides, or polysaccharides. Monosaccharides A monomer of a carbohydrate is called a monosaccharide (MAHN-oh-SAK-uh-RIED). A monosaccharideâor simple sugarâ contains carbon, hydrogen, and oxygen in a ratio of 1:2:1. The gen- eral formula for a monosaccharide is written as (CH2O)n, where n is any whole number from 3 to 8. For example, a six-carbon mono- saccharide, (CH2O)6, would have the formula C6H12O6. The most common monosaccharides are glucose, fructose, and galactose, as shown in Figure 3-6. Glucose is a main source of energy for cells. Fructose is found in fruits and is the sweetest of the monosaccharides. Galactose is found in milk. Notice in Figure 3-6 that glucose, fructose, and galactose have the same molecular formula, C6H12O6, but differing structures. The different structures determine the slightly different properties of the three compounds. Compounds like these sugars, with a single chemical formula but different structural forms, are called isomers (IE-soh-muhrz). SECTION 2 OBJECTIVES â Distinguish between monosaccharides, disaccharides, and polysaccharides. â Explain the relationship between amino acids and protein structure. â Describe the induced fit model of enzyme action. â Compare the structure and function of each of the different types of lipids. â Compare the nucleic acids DNA and RNA. VOCABULARY carbohydrate monosaccharide disaccharide polysaccharide protein amino acid peptide bond polypeptide enzyme substrate active site lipid fatty acid phospholipid wax steroid nucleic acid deoxyribonucleic acid (DNA) ribonucleic acid (RNA) nucleotide C HO H C H OH C OH H C CH2OH H C H OH O Glucose C OH C O H OH C OH H CH2OH C H CH2OH Fructose C H HO C OH H C OH H C CH2OH H C H OH O Galactose Glucose, fructose, and galactose have the same chemical formula, but their structural differences result in different properties among the three compounds. FIGURE 3-6 Copyright Š by Holt, Rinehart and Winston. All rights reserved. 56 CHAPTER 3 Disaccharides and Polysaccharides In living things, two monosaccharides can combine in a condensa- tion reaction to form a double sugar, or disaccharide (die-SAK-e-RIED). For example in Figure 3-4, the monosaccharides fructose and glu- cose can combine to form the disaccharide sucrose. A polysaccharide is a complex molecule composed of three or more monosaccharides. Animals store glucose in the form of the polysaccharide glycogen. Glycogen consists of hundreds of glucose molecules strung together in a highly branched chain. Much of the glucose that comes from food is ultimately stored in your liver and muscles as glycogen and is ready to be used for quick energy. Plants store glucose molecules in the form of the polysaccha- ride starch. Starch molecules have two basic formsâhighly branched chains that are similar to glycogen and long, coiled, unbranched chains. Plants also make a large polysaccharide called cellulose. Cellulose, which gives strength and rigidity to plant cells, makes up about 50 percent of wood. In a single cellu- lose molecule, thousands of glucose monomers are linked in long, straight chains. These chains tend to form hydrogen bonds with each other. The resulting structure is strong and can be broken down by hydrolysis only under certain conditions. PROTEINS Proteins are organic compounds composed mainly of carbon, hydrogen, oxygen, and nitrogen. Like most of the other biological macromolecules, proteins are formed from the linkage of monomers called amino acids. Hair and horns, as shown in Figure 3-7a, are made mostly of proteins, as are skin, muscles and many biological catalysts (enzymes). Amino Acids There are 20 different amino acids, and all share a basic structure. As Figure 3-7b shows, each amino acid contains a central carbon atom covalently bonded to four other atoms or functional groups. A single hydrogen atom, highlighted in blue in the illustration, bonds at one site. A carboxyl group, âCOOH, highlighted in green, bonds at a second site. An amino group, âNH2, highlighted in yel- low, bonds at a third site. A side chain called the R group, high- lighted in red, bonds at the fourth site. The main difference among the different amino acids is in their R groups. The R group can be complex or it can be simple, such as the CH3 group shown in the amino acid alanine in Figure 3-7b. The differences among the amino acid R groups gives different proteins very different shapes. The different shapes allow pro- teins to carry out many different activities in living things. Amino acids are commonly shown in a simplified way such as balls, as shown in Figure 3-7c. (a) Many structures, such as hair and horns are made of proteins. (b) Proteins are made up of amino acids. Amino acids differ only in the type of R group (shown in red) they carry. Polar R groups can dissolve in water, but nonpolar R groups cannot. (c) Amino acids have complex structures, so, in this and other textbooks, they are often simplified into balls. FIGURE 3-7 (b) Alanine (an amino acid) (c) Simplified version of amino acid CH3 H N OH C C H O H (a) Copyright Š by Holt, Rinehart and Winston. All rights reserved. BIOCHEMISTRY 57 H H N C C OH H O H CH3 H2O Glycine Alanine H N OH C C H O H H H N C C H O H CH3 N OH C C H O H (a) (b) (a) The peptide bond (shaded blue) that binds amino acids together to form a polypeptide results from a condensation reaction that produces water. (b) Poly- peptides are commonly shown as a string of balls in this textbook and elsewhere. Each ball represents an amino acid. FIGURE 3-8 Substrate Products Enzyme 1 2 3 In the induced fit model of enzyme action, the enzyme can attach only to a substrate (reactant) with a specific shape. The enzyme then changes and reduces the activation energy of the reaction so reactants can become products. The enzyme is unchanged and is available to be used again. 3 2 1 FIGURE 3-9 Dipeptides and Polypeptides Figure 3-8a shows how two amino acids bond to form a dipeptide (die-PEP-TIED). In this condensation reaction, the two amino acids form a covalent bond, called a peptide bond (shaded in blue in Figure 3-8a) and release a water molecule. Amino acids often form very long chains called polypeptides (PAHL-i-PEP-TIEDZ). Proteins are composed of one or more polypep- tides. Some proteins are very large molecules, containing hun- dreds of amino acids. Often, these long proteins are bent and folded upon themselves as a result of interactionsâsuch as hydrogen bondingâbetween individual amino acids. Protein shape can also be influenced by conditions such as temperature and the type of solvent in which a protein is dissolved. For exam- ple, cooking an egg changes the shape of proteins in the egg white. The firm, opaque result is very different from the initial clear, runny material. Enzymes EnzymesâRNA or protein molecules that act as biological catalystsâare essential for the functioning of any cell. Many enzymes are proteins. Figure 3-9 shows an induced fit model of enzyme action. Enzyme reactions depend on a physical fit between the enzyme molecule and its specific substrate, the reactant being catalyzed. Notice that the enzyme has folds, or an active site, with a shape that allows the substrate to fit into the active site. An enzyme acts only on a specific substrate because only that substrate fits into its active site. The linkage of the enzyme and substrate causes a slight change in the enzymeâs shape. The change in the enzymeâs shape weakens some chemical bonds in the substrate, which is one way that enzymes reduce activation energy, the energy needed to start the reaction. After the reaction, the enzyme releases the products. Like any catalyst, the enzyme itself is unchanged, so it can be used many times. An enzyme may not work if its environment is changed. For example, change in temperature or pH can cause a change in the shape of the enzyme or the substrate. If such a change happens, the reaction that the enzyme would have catalyzed cannot occur.
Here's how scientists figured out the age of the universe It took some cosmic detective work. by Passant Rabie Oct. 20, 2021 You never ask a cosmic being its age. But if that cosmic being encompasses all of space, time, and matter, you could get a little curious. Scientists have long been curious about the age of the universe and how much time has elapsed since the Big Bang. Today, scientists estimated the age of the universe to be approximately 13.8 billion years old. But how did scientists estimate how old the universe is, and are they sure of that number? It all comes down to ancient stars and the ever-expanding cosmos. How do astronomers calculate the age of the universe? To estimate the age of the universe, scientists rely on two main methods. Calculating the expansion rate of the universe Determining the ages of the oldest stars The Hubble Constant: Since its conception, the universe has been expanding at an accelerating rate. The universeâs expansion rate is known as the Hubble Constant, which is estimated at 46,200 mph per million light-years. The Hubble Constant was first calculated in the 1920s by American astronomer Edwin Hubble after discovering that several galaxies were moving away from Earth. Scientists looked to distant galaxies to measure how fast the universe was expanding. Hubble also noted that the further a galaxy was, the faster it was moving away. Based on Hubbleâs observations, the astronomer came up with Hubbleâs law which showed a correlation between how far an object is and the speed at which itâs receding. Using Hubble law, scientists were able to estimate the expansion rate of the universe. Scientists were then able to use the Hubble Constant to estimate the age of the universe by working backward, all the way back to the Big Bang. This extrapolation depends on the current density and composition of the universe, which shows the history of its expansion. In 2012 NASAâs Wilkinson Microwave Anisotropy Probe used that data to estimate the universe's age to be 13.772 billion years old, give or take 59 million years. A year later, The European Space Agencyâs Planck spacecraft estimated the universe's age to be 13.82 billion years. Ancestral stars: Another way to determine the age of the universe is to look to the oldest stars. The universe canât be younger than its oldest stars. Therefore, to narrow down the age of the universe, scientists measure the ages of the very first stars that formed in the cosmos. The lifecycle of a star depends on its mass, with high mass stars burning fuel at a faster rate and therefore dying out faster while low mass stars can live up to 20 billion years. Globular clusters are a dense stellar collection of around a million stars which all formed roughly around the same time. These clusters can then serve as timekeepers for the universe. By determining the masses of their stars, scientists can estimate when the globular cluster formed. The oldest globular clusters contain stars that are 0.7 times less massive than the Sun, which suggests that they are between 11 to 18 billion years old. What came before the Big Bang? Scientists can trace the universe back to its explosive birth, the Big Bang. But what happened before this theoretical birth of the cosmos? The universe may have been a singularity, all compact within a form that is smaller than a subatomic particle. Itâs difficult to imagine what caused this matter to exist, but one theory even suggests that our universe was born from another universe while another imagines a series of universes being born out of one another like a formation of bubbles. Meanwhile, another theory suggests that the universe goes through an endless cycle of death and rebirth, born from its own demise. How old is the universe in seconds? If the universe is indeed cyclical, then time becomes irrelevant. But just in case youâre still attached to the modern way in which we measure the progression of life, then the age of the universe comes up to about 436,117,076,900,000,000 seconds.
3.2 Sexual relationships: Christian teachings about sexual relationships Christianity teaches that sexual relationships should only take place between a man and a woman who are married to each other. The Bible says throughout that: Sex should only take place within marriage and that all other forms of sexual activity are forbidden Marriage should be respected, and the sexual relationship between a husband and wife should be kept pure After death, God will judge and punish people who commit adultery and those who are sexually immoral. Christian attitudes towards sexual relationships can vary. Some Christians agree with these points. They think sex before marriage, adultery, prostitution, and homosexual relationships are wrong, as highlighted in 1 Corinthians 6:7â20, for example. They think these kinds of sexual relationships disrespect the human body which, according to the Bible, is the âtemple of the Holy Spiritâ and belongs to God. The purity ring (Figure 3.2) is how some Christians choose to show their views on sexual purity. Two Christian views on sexual relations View 1: Some Christians argue the main purpose of sex within marriage is to have children. This means sex should only take place between a man and a woman, as same-sex couples cannot naturally produce a child. Christians base this belief on a biblical command from the Creation in Genesis 1:28. This says human beings are to be âfruitful and increase in numberâ. Other Christians say that sex between a husband and wife has another purpose: it is an important expression of their love and unites them as a couple. This comes from Genesis 2:24, which refers to a man and a woman being united in âone fleshâ. View 2: Some Christians think the attitudes in View 1 are outdated. They would argue that Christianity is about agape (selfless, unconditional love). They think it is wrong to criticise sex in a relationship that is committed and loving, even if those involved are cohabiting rather than married. This means some Christians might also find sexual relationships between homosexuals acceptable within a permanent and caring relationship. Christians disagree with people having sex within casual or short-term relationships. Atheist and Humanist attitudes towards sexual relationships Depending on their personal views, atheists may have differing opinions about: Some sexual activities Sex outside of marriage Adultery Prostitution Homosexuality. Their views would not be based on religious teachings. Freedom and choice are two important Humanist values, providing they do not cause harm to anyone else. Humanists believe consenting adults should be free to have the sexual relationships they want as long as those relationships do not damage others. Humanists are opposed to all forms of sexual abuse and exploitation. As with any choice made by human beings, Humanists believe people should think about the consequences of their sexual relationships, making responsible and thoughtful choices that take into account the happiness of those involved