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course phy 201
1/21 12:08
Question: `q001. There are two parts to this problem. Reason them out using common sense.
If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark?
Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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Your solution:
In this equation you divide 2 by 20 and 30 giving you 10-15 seconds. Then you multiply 2 by 7 giving you 14 and add 14 to 10, this gives you 24.
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Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds.
It then repeats the process, this time passing the milepost at a speed of 20 mph. This time:
• Will the vehicle require more or less than 10 seconds to reach the lamppost?
• Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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Your solution: The second time it will take less than 10 seconds to reach the sign because the car is going at a faster rate. For the second question, it will change by less since the speed is still increasing by the same rate.
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Question: `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second.
We wish to compare the rates at which two different automobiles increase their speed:
Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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Your solution:
The first automobile seed up at a faster rate. We find this be subtracting the final minus initial speed of the car then diving that number by the seconds at which it sped up.
Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile.
Which team will win and why?
If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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Your solution: The second team would win. We find the answer by dividing Newtons by kg in each problem. He second team gives us a larger number of 2.5 compared to 2. Still, if the first team pulled with 500N of force they would lose because the answer is less than 2.5.
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Question: `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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Your solution: Person number two would knock over player 1. We can find this by multiplying the wt by the seconds. Player two gives us a larger number of 4000.
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Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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Your solution: The second person generates a larger number if we divide the oz by wt. Climber two has a very slight advantage at .007 more than climber 1.
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Question: `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop.
Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long?
Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great?
Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?
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Your solution: The second car would take double the time as the first car to stop. For average coasting velocity it’s the same as above, it would take double the time. However, the distance traveled by the faster car will be more than twice that of the slower car.
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Question: `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length.
Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?
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Your solution: I think it would be more than 7 feet because I would assume that the rope would stretch more between 100 and 125 lbs.
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Question: `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet.
The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force).
When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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Your solution: The skate would go more than twice as far. In this problem you have to take into account force and distance. Both are double making it 4 times as far.
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Question: `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet.
To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first?
To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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Your solution: The larger sphere will appear dimmer because the light in the glass will have to travel a greater distance to get out of the frosted glass. For the second question the answer would be the same appearing to have less than half the brightness of the first.
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Question: `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius.
Place the following in order, from the one requiring the least energy to the one requiring the most:
Increasing the temperature of the ice by 20 degrees to reach its melting point.
Melting the ice at its melting point.
Increasing the temperature of the water by 20 degrees after all the ice melted.
At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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Your solution: Increasing the temperature of the water by 20 degrees after all the ice melted.
Increasing the temperature of the ice by 20 degrees to reach its melting point.
Melting the ice at its melting point.
I would think it would tae more effort for the state change to be cooled gain than heated to evaporate.
I would conclude that ice melts at 0 degrees Celsius. We can see this in the above equation when it states that the s=ice cube is first melted a little bit at 0 degrees Celsius.
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Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards, one at either end of the pool, are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you.
Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much.
If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point?
How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?
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Your solution: At the center point I would expect that since both are coming at the same time that it would be double at high at 12 inches. I honestly don’t quite understand how to reach the answer to the second question.
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