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02-OUR WORLD free text
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Of the 7 billion people on Earth roughly 0:02 6 billion own a cell phone which is 0:05 pretty shocking given that only 4 and2 0:07 billion have access to a working toilet 0:09 so how are these popular gadgets 0:11 changing your body and brain If you're 0:13 looking down at your phone right now 0:15 your spine angle is equivalent to that 0:17 of an 8-year-old child sitting on your 0:19 neck which is fairly significant 0:21 considering people spend an average of 0:23 4.7 hours a day looking at their phone 0:26 this combined with the length of time 0:28 spent in front of computers has led to 0:30 an increase in the prevalence of myopia 0:32 or nearsightedness in North America in 0:34 the 1970s about one quar of the 0:36 population had myopia where today nearly 0:39 half do and in some parts of Asia 80 to 0:41 90% of the population is now nearsighted 0:44 and it can be hard to put your phone 0:46 down take for example the game Candy 0:48 Crush as you play the game you achieve 0:50 small goals causing your brain to be 0:52 rewarded with little bursts of dopamine 0:54 and eventually you're rewarded in the 0:56 game with new content this novelty also 0:58 gives little bursts of dopamine and 1:00 together create what is known as a 1:01 compulsion Loop which just happens to be 1:04 the same Loop responsible for the 1:05 behaviors associated with nicotine or 1:07 cocaine our brains are hardwired to make 1:10 us novelty seeking and this is why apps 1:12 on our phones are designed to constantly 1:14 provide us with new content making them 1:16 hard to put down as a result 93% of 1:19 young people aged 18 to 29 report using 1:21 their smartphone as a tool to avoid 1:23 boredom as opposed to other activities 1:26 like reading a book or engaging with 1:27 people around them this has created a 1:29 new term nomophobia the fear or anxiety 1:32 of being without your phone we also see 1:35 a change in brain patterns Alpha rhythms 1:37 are commonly associated with wakeful 1:39 relaxation like when your mind wanders 1:41 off whereas gamma waves are associated 1:44 with conscious attentiveness and 1:46 experiments have shown that when a cell 1:47 phone is transmitting say during a phone 1:49 call the power of these Alpha Waves is 1:52 significantly boosted meaning phone 1:54 Transmissions can literally change the 1:56 way your brain functions your smartphone 1:58 can also disrupt your sleep the screen 2:00 emits a blue light which has been shown 2:02 to alter our circadian rhythms 2:03 diminishing the time spent in deep Sleep 2:06 which is linked to the development of 2:07 diabetes cancer and obesity Studies have 2:10 shown that people who read on their 2:11 smartphone at night have a harder time 2:13 falling asleep and produce less 2:15 melatonin a hormone responsible for the 2:17 regulation of sleep wake Cycles Harvard 2:20 medical school advises the last 2 to 3 2:22 hours before bed be technology free so 2:24 pick up a book before bed instead of 2:26 course smartphones also completely 2:28 change our ability to access information 2:30 most notably in poor and minority 2:32 populations 7% of Americans are entirely 2:35 dependent on smartphones for their 2:37 access to the internet a 2014 study 2:40 found that the majority of smartphone 2:41 owners use their phone for online 2:43 banking to look up medical information 2:45 and searching for jobs so while phones 2:47 are in no way exclusively bad and have 2:50 been part of a positive change in the 2:51 world there's no denying that they are 2:53 changing us but many successful people 2:56 have now decided to take smartphone 2:58 vacations in order to increase 3:00 productivity in our new ASAP thought 3:01 video we break down the top six reasons 3:04 you should take a smartphone vacation 3:06 and how it could benefit your life right 3:08 now and subscribe for more weekly 3:09 science videosIntroduction 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.02-OUR WORLD EN-GR02-OUR WORLD scrambled02-OUR WORLD GR-ENMake a multiple choice quiz for my year 8 science students based on the science in this transcript from a video: 3°C 0:04 It can be the difference between snow and sleet 0:08 Wearing a jacket or not 0:11 In your day-to-day life, it may not seem significant 0:15 But 3°C of global warming would be catastrophic 0:20 Heatwaves, droughts, extreme precipitation, even fire 0:25 3°C of warming is really disastrous 0:28 The scary thing is, the world is well on its way there 0:32 Since the industrial revolution, the Earth has warmed between 1.1°C and 1.3°C 0:40 This is a problem that babies you pass in the street will have to live with 0:46 Children born today... 0:47 ...are up to seven times more likely to face extreme weather than their grandparents 0:52 If global temperatures do rise by 3°C... 0:55 ...what would their world look like? Climate change is already having devastating effects 1:03 Rising sea levels 1:05 Desertification 1:07 Hollywood has always enjoyed imagining the end of the world 1:11 While blockbusters like this are clearly fiction... 1:14 ...this film will show the scenario we all face... 1:17 ...unless more drastic measures are taken to stop burning fossil fuels 1:30 In some parts of the world the effects of inaction are already clear 1:35 The slums of Bangladesh’s capital are filling up with climate migrants 1:41 Minara comes from Bhola District, an area in southern Bangladesh 1:46 There, like many other parts of the country... 1:49 ...rivers swollen by heavier rain and melting Himalayan glaciers... 1:53 ...are washing away people’s homes 1:56 Many, like her, have lost everything 2:00 Our home in Bhola had endless amounts of land 2:03 There was lots of space for farming, we had a spacious house 2:08 There were different types of fruits, vegetation and trees growing at home 2:12 We used to eat the fruit from our own trees 2:18 I can’t eat them now because they don't exist anymore 2:21 Since the river flooded for the third time, I had to flee to Dhaka 2:26 Life was much better back home 2:29 It was unbearable to live through, truly intolerable 2:33 We didn’t have the time to save anything at all 2:38 1.1°C to 1.3°C of global warming has already transformed Minara’s life 2:45 It’s one of the reasons why so many migrants like her... 2:47 ...are moving to the city each year... 2:50 ...nearly 400,000 according to the last estimate 2:53 And climate models show there could be much worse to come How climate modelling works 3:02 Climate scientist Joeri Rogelj... 3:04 ...has spent the last ten years modelling future climate scenarios... 3:08 ...for the United Nations 3:10 The models we use to carry out this exercise... 3:13 ...really represent the state of the art... 3:15 ...of our current knowledge of climate change and where we are heading 3:19 Joeri’s projections use data collected by hundreds of scientists around the world 3:26 Here this is the 3°C level... 3:28 ...and so there is at least a one-in-four chance that under current policies... 3:32 ...we would hit 3°C by the end of the century 3:36 This is just one of the scenarios Joeri looks at 3:40 Another one imagines that all policy promises are kept 3:44 The most optimistic assumes that all promises have been kept... 3:47 ...and net-zero targets are met 3:50 Where our best estimate ends up around 2°C at the end of the century... 3:54 ...there is still a one-in-20 chance that we end up with 3°C instead 3:59 One would not be entering a plane if there is a one-in-20 chance... 4:03 ...that the plane will crash Nowhere is safe from global warming 4:07 A rise of 3°C would affect everyone 4:10 Even wealthy cities in rich countries wouldn’t be immune to the consequences 4:15 European capitals like Paris and Berlin... 4:18 ...would bake under more extreme heatwaves 4:22 Frequent storm-surges in New York could turn parts of the city desolate 4:27 In many ways, cities magnify, intensify climate events 4:33 Cities are hotter than the places around them... 4:36 ...they tend to be more vulnerable to flooding 4:39 And you can get a really bad event in a city in a way that you can’t in the countryside 4:46 And because of their denser populations... 4:49 ...disasters in a city affect far more people 4:52 Some cities might be badly prepared for the changes coming 4:56 But they have the means to adapt 4:59 Cities tend to be wealthier than surrounding places 5:03 They have a lot of amenities 5:05 A city that has taken seriously the risks of a 3°C world... 5:08 …wouldn’t necessarily be a worse place to be in a 3°C world 5:12 But a city that hasn’t prepared for these sort of eventualities... 5:16 ...that might be a really nasty place The impact of prolonged droughts 5:20 So far, many developed cities have got off lightly... 5:24 ...but some rural parts of the world are suffering disproportionately 5:29 Smallholders—small-scale farmers—are particularly vulnerable to climate change 5:35 And there are over 600 million around the world 5:38 Smallholders with farms under two hectares... 5:40 ...produce around a third of the global food supply 5:46 Central America’s “Dry Corridor”... 5:48 ...supports a mix of smallholdings and medium-sized farms 5:53 Sandwiched between the Pacific Ocean and the Caribbean Sea... 5:56 ...the area is prone to droughts 6:08 Israel Ramírez Rivera is a smallholder in Guatemala 6:12 Here, climate change is making the dry seasons longer, and more severe 6:18 This is the biggest ear of maize that this plot could deliver 6:23 He depends on his crops of corn and beans 6:26 But they’re getting harder to grow 6:30 The surrounding mountains... 6:32 ...used to provide us with native food... 6:38 ...and now that isn’t an option anymore... 6:41 ...due to climate change and its effects 6:46 Nearly two-thirds of the smallholders in the Dry Corridor now live in poverty 6:52 The impact of all of this for us... 6:59 ...malnutrition among children 7:03 We’ve lost a few 7:07 For my crops especially, the midsummer heat is harder than before 7:16 The plant dries up and can’t provide us... 7:19 ...with the necessary food provision 7:24 Severe droughts in Central America... 7:26 ...are now four times more likely than they were last century 7:30 Many families from here have gone to the States 7:37 The economic despair and debts... 7:44 ...have pushed many people from this community to do this journey 7:53 Migration from Guatemala to the United States has quadrupled since 1990 7:59 Not all of this has been due to climate change 8:02 But longer droughts would force even more to move 8:05 In a 3°C world, annual rainfall in this region... 8:09 ...could drop by up to 14% 8:12 At 3°C, over a quarter of the world’s population... 8:16 ...could endure extreme droughts for at least a month of the year 8:19 Northern Africa could see droughts that last for years at a time Rising sea levels, storm surges and flooding 8:24 But for some, too much water will be the problem 8:29 10% of the world’s population lives on a coastline... 8:32 ...that’s less than 10 metres above sea level 8:35 For these coastal inhabitants, a 3°C world would spell disaster 8:40 By 2100, global sea levels could have climbed by half a metre from 2005 levels 8:46 Low-lying cities like Lagos would be especially vulnerable... 8:49 ...with up to up to a third of the population displaced 8:54 And in Fiji, rising waters are already upending lives 9:04 You can see the graveyard there, it’s all under water now... 9:08 ...due to this rising sea level and climate change 9:15 The village of Togoru in Fiji is being swallowed by the sea 9:19 Barney Dunn, the village headman, has seen over half the village disappear 9:24 Relatives’ houses have been abandoned, and family graves are now under water 9:29 We have been asked by the government to relocate... 9:32 ...but no one wants to relocate... 9:34 ...because we have our great-great-grandparents down there in the sea 9:39 This is the place we’ve been brought up in 9:41 ...it’s not easy to leave 9:44 Past attempts to build a seawall haven’t worked 9:48 But Barney sees building a new one as the village’s only hope 9:52 If they do that, maybe we can save whatever is left 9:56 But if we don’t have the seawall, then it will be keep eroding and time will come... 10:01 ...maybe in ten,15 years, Togoru will be all eroded 10:05 Rising seas also mean storms cause more floods 10:11 And many more countries could suffer 10:14 The Philippines and Myanmar are just two countries... 10:17 ...that will also see an increase in storm surges in a 3°C world 10:21 To escape, many will move… 10:24 …often, to urban areas Extreme heat and wet-bulb temperatures 10:27 Half the world’s population already lives in cities... 10:31 ...almost a third in slums 10:36 For them, a 3°C world could be deadly 10:40 Minara has moved to Dhaka to escape the impact of climate change 10:44 But life could get even worse for her 10:47 I’m struggling a lot nowadays 10:49 The heat during the day is unbearable 10:52 Even late at night it doesn’t cool down 10:57 The heat is getting more intense every day 10:59 I mean, it’s going to get much worse 11:03 I can barely survive it now, how will I live through it in the future? 11:08 Dhaka is getting hotter 11:11 In the last 20 years the average daytime temperature... 11:13 ...has crept up by nearly half a degree 11:17 Days that approach 40°C are now being reported 11:20 And high so-called wet-bulb temperatures are on the rise 11:26 A wet-bulb temperature is a measure of heat and humidity 11:30 Humans cool themselves by sweating… 11:32 But in these conditions, when relative humidity is near 100%... 11:36 ...sweat doesn’t evaporate well 11:38 So people can’t cool down… 11:41 ...even if given unlimited shade and water 11:45 At a high wet-bulb temperature, the body can’t lose heat... 11:49 ...and so it gets hotter and hotter... 11:51 ...and the body is designed to work at a given temperature 11:53 And if it gets too hot inside, you will die 11:58 The human limit for wet-bulb temperatures is 35°C... 12:02 ...around skin temperature 12:04 Dhaka will have a much higher chance... 12:05 ...of reaching dangerous wet-bulb temperatures... 12:07 ...if global warming reaches 3°C 12:12 You can’t really adapt to that 12:14 You have to get out. If the temperature is so high that you can’t work... 12:20 ...can’t do hard manual labour outside for significant parts of the year... 12:25 ...then many places will become functionally no longer part of the economy 12:33 Jacobabad in Pakistan, and Ras al Khaimah, in the United Arab Emirates... 12:37 ...have already recorded deadly wet-bulb temperatures 12:40 More of the tropics and the Persian Gulf... 12:43 ...as well as parts of Mexico and the south-eastern United States... 12:47 ...could all get to this threshold by the end of the century 12:50 Climate modelling might show us the weather Increased migration and conflict 12:52 But it doesn’t show us its other effects on society 12:56 Established migration patterns could change 12:59 Climate disasters may exacerbate reasons people cross borders 13:03 Within countries, more people will move to cities 13:07 In a 3°C world, tens of millions of people a year... 13:10 ...could be displaced by disasters made worse by climate change 13:15 When people are displaced by climate... 13:18 …they may well go to cities... 13:19 ...because cities are the places that attract people from the countryside already 13:25 A lot of people who can get to the developed world... 13:28 ...not least because the developed world tends to be less hot, will give that a go 13:35 As migration around the world increases... 13:38 ...there could be more competition for fewer resources 13:42 Water—already a highly contested resource—will be a focal point 13:47 Turkey’s new Ilisu dam has reduced the flow of water into Iraq 13:53 China lays claim to rivers vital to India and Pakistan 13:57 The prospect of a water-conflict makes people very uneasy 14:03 How national tensions would exacerbate those sorts of reactions... 14:08 ...in a 3°C world... 14:09 ...is the sort of thing that no one should really want to find out 14:14 I think you’d have to be incredibly sanguine... 14:16 ...not to think that the sort of climate extremes that we talk about... 14:19 ...in a 3°C world wouldn’t lead some places... 14:22 ...to the brink of societal collapse 14:25 Those lucky enough to escape unrest... Adaptation and mitigation are crucial 14:28 ...would still have to adapt to a radically different world 14:32 People can adapt to climate change in all sorts of ways, one of the most obvious ones... 14:37 ...is air conditioning 14:39 But other ways to adapt at a local or regional level... 14:42 ...I mean, one of the most obvious is diversifying agriculture 14:47 There are physical things you can do, like seawalls 14:52 The fact that people can adapt and that adaptation will reduce suffering... 14:57 ...doesn’t mean that it will eliminate suffering 15:00 Suffering is built into this whole process of heating up the planet 15:06 Adaptation will only get the world so far 15:09 The best way to deal with a 3°C world... 15:12 ...is not to go to a 3°C world 15:14 And that’s why increasing efforts on mitigation are important 15:17 It’s why working towards negative emissions... 15:20 ...that could bring down the temperature after it peaks are important 15:25 Once you get to a 3°C world, you are in real bad global trouble 15:33 The scale of change needed... 15:35 ...and the slow progress of governments so far... 15:38 ...means 3°C of warming is uncomfortably likely unless more is done 15:44 Despite existing pledges, greenhouse-gas emissions... 15:48 ...are still set to rise by 16% from 2010 levels by 2030 15:54 The need to act has never been clearer 15:57 There’s still time to reduce emissions, so that a 3°C world remains fiction... 16:02 ...rather than becoming factIn this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level02- Identifying Parts of a Word