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course Phy 241
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: Part 1) 30mph - 20 mph will find the change in mph. (10pmh) divide that by two to find out how many seconds, since 2 pmh increase every second. 5 seconds.
Part 2) multiply our time by 2; 7 * 2 = 14. There is a 14 mph increase. Add 14 to 10 and you will get the answer: 24 mph after 7 more seconds.
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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: 1) The vehicle will requie less time to reach the lamppost because it is going faster initially. 2) At the same rate, the mph would increase by the 10 mph. The rate would have to be ajdusted for the vehicle because it would not take it 10 seconds to get to the pole, going 10 miles per hour faster than our initial vehicle.
`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: we need to find the change in speed in both instances then divide by the time it took to reach the change.
1) 30 - 20 = 10; 10/ 5 = 2 mph/sec
2) 90 - 40 = 50; 50/ 20 = 5 / 2 = 2.5 mph/sec
The vehicle going from 40 to 90 in 20 seconds speeds up at a greater rate than the vehicle going from 20 to 30 in five seconds .
`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?
Your solution: 3000 Newtons / 1500 kg = 2 Newtons per kg for team one; team two would be 5000 Newtons / 2000 kg = 2.5 Newtons per kg. The second team will win because there are 2.5 Newtons per 1 kg pulling which is more than team ones 2 Newtons per 1kg pulling.
5000-500 = 4500 Newtons / 2000 kg = 2.25 Newtons / kg which is stil more pull than the first team.
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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?
Your solution: 250 lb / 10 feet per second = 25 lb / feet per second for the first boy and 200 lb / 20 feet per second = 10 lb/ feeet per second. The 200 lb boy would be pushed backwards immediately bc the force of the first boy is far greater than his.
Feedback: I did this equation opposite of how I should have, instead of dividing to find the force i should have multiplied, making the outcome opposite, having the second boy dominating by 1500 lb*fps.
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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?
Your solution: we first calculate the ounces / per for eat climber. 10 oz / 150 lb = 0.067 oz/lb for the first climber and 12 oz / 200 lb = 0.06 for the seconds. The second climber, therefore, has more energy.
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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.
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Your solution:
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?
The second vehicle, going twice as fast, will take longer to come to a stop. I think it will take less than twice as long to stop with gravity and other resistance
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?
The vehicle going faster will have a greater coasting velocity bc it is initially going faster than the other vehicle. I believe it will also be a little less than twice that of the slower vehicle.
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?
The distance traveled by the faster car will be a little less than twice that of the slower car. I would say that by having a greater coasting distance you would have to have a greater total distance.
FeedBack: Would not have guess it went four times as far as the other vehicle but i would calculate in the wind resistance.
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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: First you note that from 100 to 150 the rope strecthes 4 feet. between 150 and 200 it only increases its stretch by 3 feet. This would lead us to believe that with less weight, theres more strecth. so a 125 lb person would strecth the rope a bit more than 7 feet.
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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?
Your solution:
She will be expected to travel twice as far as she did by being pulled back 4 feet. But also the force is doubled. So she would possible go four times as far as she would have intitally.
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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: 1) I would think it would be dimmer because the distance between the bulb and the glass is larger than the first.
2) I would say half the brightness bc it's twice the distance.
Feedback: 1) the solution states that rom a distance they both look the same as far as brightness bc both bulbs emmit the same amount of light through the same amount of frosting on the glass. 2) The brightness would be 1/4 the that of the smaller sphere, not because of distance between bulb and sphere but bc of surface area increased. Less light is illuminating on a square inch of the larger sphere than of the smaller sphere.
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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:
Melting the ice at its melting point.
Increasing the temperature of the ice by 20 degrees to reach its melting point.
Increasing the temperature of the water by 20 degrees after all the ice melted.
Ice melts and freezes at 0 degrees celsius. The temperature remains constant until all ice is melted.
`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:
The peaks are six inches high and both peaks are meeting since you in the middle, displacement is 12 in. The waves are six feet apart so if you move six feet you will be getting both peaks again. I would think if you moved three feet you would be in the middle of one wave and catching the middle of the next wave also since itd also be three feet into the next peak. To catch a peah and a valley you'd have to be at 1.5 or 4.5 feet in either direction.
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Question: `q013. This problem includes some questions that are fairly straightfoward, some that involve more complicated considerations, and possibly some that can't be answered without additional information.
We're hoping for some correct answers, but we expect that few students coming into this course will be able to think correctly through every nuance of the more complex situations. On these questions we are hoping for your best thinking without being particularly concerned with the final answer.
A steel ball and a wood ball are both thrown upward and, between release and coming to rest at maximum height, both rise with the same average speed. If not for air resistance they would both come to rest at the same time, at the same height. However air resistance causes the wood ball to stop rising more quickly than the steel ball.
Each ball, having risen to its maximum height, then falls back to the ground.
***Assuming the steel ball weighs more than the wooden ball:***
Which ball would you expect to have the greater average velocity as it falls?
I would think the steel ball would have greater average velocity, falling from a higher point, with less wind resistance.
Which ball would you expect to spend the greater time falling?
The wood ball, the air resistance falling down would slow it some, and the density of the wood ball would be less than that of the steel ball.
Which ball would you expect to hit the ground first?
I would expect the balls to hit at the same time. We assume it has longer to fall but we also threw the balls at the same time and the air resistance on the wooden ball caused it to not rise as high as the steel ball. I feel like it would even back out in the fall as it unevened out in the rise.
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Question: `q014. If you double the voltage across a certain circuit you double the current passing through it. The power required to maintain the circuit is equal to the product of the current and the voltage. How many times as much power is required if the voltage is doubled?
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Your solution: double voltage * double current = quadruple the power required to maintain the circuit.
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