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course Phy 241
06/09/2013 11:30 pm
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:
It will take 5 seconds for the speedometer to move from 20 mph to the 30 mph mark. If the speedometer initially reads 10 mph, then it should read 24 mph, after 7 seconds.
a. 30 mph - 20 mph = 10 mph
10 mph/ 2 mph/sec = 5 seconds
b. 2 mph/sec * 7 seconds = 14 mph
14 mph + 10 mph = 24 mph
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Self-critique Rating: 3
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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 vehicle will require less than 10 second to reach the lamppost because the automobile is moving down a hill, which means the velocity is increasing at the same rate every second.
Which implies less time on the second run, the key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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Self-critique Rating: 2
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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 rate of change of the first automobile's speed changes from 20 mph to 30mph is 2 mph per second.
30 mph - 20 mph = 10 mph
10 mph/5 seconds = 2 mph/ sec
The rate of change of the second automobile's speed changes from 40 mph to 90 mph is 2.5 mph per second.
90 mph - 40 mph = 50 mph
50 mph/20 seconds = 2.5 mph/seconds
The second automobile increased its velocity at a rate of .5 mph / second, which is greater than that of the first.
2.5 mph/ seconds - 2 mph/seconds = .5 mph/seconds
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Self-critique Rating: 3
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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 team that will win is the second team since they can increases velocity more quickly and will stay ahead.
First team Second team
3000 N /1500 kg 5000 N/2000 kg
= 2 N/kg = 2.5 N/kg
The other team would still lose because the second team produces a rate of 2.25 Newtons per kg, which is still more than the 2 Newtons / kg of the first team.
5000 N - 500 N = 4500 N
4500 N/2000 kg = 2.5
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Self-critique Rating: 3
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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 predict will be moving backward immediately after the collision, and why?
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Your solution:
In order to find which player that will be moving backwards after the collision, we must multiply speed by mass to get the determining quantity, which is called momentum. From that, the second player will dominate the collision.
First player Second player
250 lb * 10 ft/sec = 2500 lb ft / sec 200 lb * 20 ft/sec = 4000 lb ft / sec
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Self-critique Rating: 3
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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:
We compared the number of ounces of Cheerios per pound of body weight, in which the first climber has .06 oz / lb of weight, while the second has.067 oz / lb. The second climber therefore has more energy per pound of body weight.
11 oz / (200 lb) = .06 oz/lb
10 oz / (150 lb) = .067 oz/lb
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Self-critique Rating: 3
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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 slower auto will take longer to stop because if you neglect air resistance, the slope is the same for both cars, which means both automobiles will change velocity at the same rate. So in this case the slower auto would require exactly twice as long.
However if you include air resistance, the faster car experiences more so it actually takes a bit less than twice as long as the slower.
As mentioned before, if air resistance is neglected, the velocity would change at a constant rate. So with that knowledge, the faster car would be exactly twice as great if air resistance is neglected.
Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far.
If there is air resistance then the velocity of the faster car decrease more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.
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Self-critique Rating: 2
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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:
So far we know that, from 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb range, each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft.
A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet. You can tell from the graph that the line would be concave downward, which shows that graph is increasing at a decreasing rate
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Self-critique Rating: 3
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 distance through which the force acts will be twice as great, in turns, would double the distance. The cause of this is the doubled pullback and the linear proportionality relationship for the force. In an addition, the average force is also twice as great, which alone would double the distance. In the end, we must double the doubling, and as a result, she will go 4 times as far.
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Self-critique Rating: 2
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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:
While both bulbs will send out the same energy per second, the surface of the second bulb will indeed be dimmer than the first. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area.
The sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.
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Self-critique Rating: 2
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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: From the facts above, the temperature in which ice melts is 0 Celsius. Its 0 Celsius because the temperature was the same when a little of the ice is melted as when most of it is melted, and melting takes place at 0 Celsius.
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Self-critique Rating: 3
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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 'fill 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 height of the two 6-inch peaks that bob up and down in the center point is 12 in.
6 in + 6 in = 12 in
The distance I have to move toward one end or the other in order for peaks to meet alleys is 24 in.
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Self-critique Rating: 3
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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 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.
Which ball would you expect to have the greater average velocity as it falls?
Which ball would you expect to spend the greater time falling?
Which ball would you expect to hit the ground first?
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Your solution:
The ball that has the greater average velocity as it falls is the steel ball. The reason why the steel ball is most likely to have a greater average velocity than the wood ball is because the steel ball lacks the amount of air resistance.
The ball that spend the greater time falling is wooden ball because its lighter than the steel ball.
The ball that will hit the ground first is the steel ball because it has a heavier mass than the wooden ball.
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Self-critique Rating: 3
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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:
If the voltage is doubled and goes through a circuit that doubles the current passing through it, the amount of times as much power is needed is 4x.
Voltage = 2x
Current = 2
Power = 2x * 2 =4
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Self-critique Rating: 3
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