qainitialasst6

course Phy 201

???T??}*???·????·assignment #006006. Physics

qa initial problems

01-28-2009

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15:03:16

`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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RESPONSE -->

It will take 5 seconds for the care to move from 20 mph to 30 mph. because there is 10 mph in between them and 10/2=5

7 seconds later the speedometer will read 24mph because you start at 10 and 7sec* 2mph=14 then add the 14 to the 10 to get 24.

confidence assessment: 3

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15:12:54

`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.

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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

The vehicle will require less than 10 seconds to reach the lamppost. If it took 10 seconds the first time to pass it at a speed of 20 mph because of the increase of 2mph every sec. for 10 sec. then the second time it would pass it would be a lot faster because of the increase in speed. The car would reach the lamppost in half the time. And the speed would be 10mph greater than before.

confidence assessment: 3

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15:59:59

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

The automobile that speeds up from 40 mph to 90 mph in 20 seconds speeds up at a greater rate. Because 20-30=10mph/5sec=2mph/sec and 40-90=50mph/20sec=2.5 mph/sec.

confidence assessment: 3

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16:13:04

4. 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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

The team that exerts a net force of 5000 Newons on a 2000 kg automobile. will be the team that will win today because they exert the most force. Because 3000Newtons/1500 kg is 2. And 5000 Newtons/2000kg =2.5. The other team would not win if a force of 500 newtons is added to the loosing teams amount it would = 3500Newtons/1500kg= 2.333 which is still less than the 2.5.

self critique assessment: 3

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16:26:25

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

I think that the 200lb player will be the one that will be moving backward immediately after the collision. Becsuse if you devide the mass by the velocity the 250lb/10ft=25lbs of force exerted on the other player verses the 200lb/20ft =10lbs of force.

confidence assessment: 2

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16:38:56

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

I think that the 200 lb climber would be able to climb further up the mountain. Because 200lb*12oz=2400 and 150lb*10oz=1500 so the larger person with the larger meal should have more energy to burn

confidence assessment: 1

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20:46:43

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

The faster automobile will take longer to stop because it is going faster. Since the automobile si moving twice as fast as the other it should take him about twice as long to stop.

confidence assessment: 2

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20:56:01

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

I think that the cord would stretch exactly 7 feet or a little less beyond its initial length.because there is 4 ft difference between 5ft and 9ft. and 4 devided by 2 . so add the 5 and 2 that gives you 7.

confidence assessment: 1

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21:17:55

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

I think that the skater will travel more than twice as far as when she was pulled back 4 feet because the force increases in direct proportion to the pullback.

confidence assessment: 1

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21:28:28

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

I think that the larger one will appear brighter from a half mile away because it fills up more space in the sphere so the light would be larger. I think that the moth walking on the surface would think that the second sphere would be half the brightness of the first because the sphere is 2 times the size of the first sphere and the futher you go from the light source the dimmer the light will become. Small area will be brighter than the larger area.

confidence assessment: 2

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21:37:54

`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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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 think the ice appears to melt at 0 degrees celsius. Because the temperature remains at 0degrees celsius after the ice has started to melt and the water and ice combination is still at 0 degrees celsius.

confidence assessment: 1

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21:45:19

`q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards 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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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

I really don't think that i have enough information to answer this question but if I had to guess. I would say that you would bob up and down half the size of the waves so 3 inches. ""And that you will have to get six inches from the edge of teh pool in order for the peaks to meet the valleys in order to place you in relatively calm water.

confidence assessment: 0

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You are doing some good thinking and have a lot of good answers.

However you also need to be self-critiquing your work in terms of the given solutions, and you don't seem to be doing so.

&#To run the program correctly you click on the Next Question/Solution button (top left), enter your response in the Answer box (the box at top right), click on the Enter Response button (above the top right box), then again on the Next Question/Solution botton (top left) to see the solution, and finally enter your response or self-critique in the Answer box and click on Enter Response.

You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc.

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&#It is recommended that you periodically review the document on responses and self-critiques located at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm, which is relevant to your responses and self-critiques on this exercise. &#