AP Physics C Summer Assignment Help

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1

<li>[SF6 1.P.035.]
A point is located in a polar coordinate system by the coordinates r = 2.0 m and = 35°. Find the x and y coordinates of this point, assuming the two coordinate systems have the same origin.</li>
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<p>For my X coordinate I got 2, which it is telling me is wrong
For my y cooridinate I got 1.12, which it is telling me is correct</p>

<p>I did:</p>

<p>x= r cos of theta = 2cos(35)= -1.80738
y= r sin of theta = 2 sin(35)= -.856368</p>

<p>r = square root of (X^2 + y^2)
r= square root of apprx. 4
so r = 2, so why is my x coordinate wrong? r IS the x coordinate. what did I do wrong.</p>

2

<p>Cos and sin are positive in the first quadrant. Try doing the math in degree mode. Secondly, the x coordinate can't be 2 because r is 2 and theta isn't 0</p>

3

<p>thanks, i changed to degree and got the right answer. thanks alot</p>

4

<p>How many nanoseconds does it take light to travel 4.80 km in vacuum?</p>

<p>I calculated the speed of light and then I still am comign up with the wrong answer. What are the steps in doing this? Thank you</p>

5

<p>Here's the answer: <a href="http://www.google.com/search?hs=IRD&hl=en&lr=&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial_s&q=4.8+km+%2F+c+in+nanoseconds&btnG=Search%5B/url%5D"&gt;http://www.google.com/search?hs=IRD&hl=en&lr=&safe=off&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial_s&q=4.8+km+%2F+c+in+nanoseconds&btnG=Search&lt;/a&gt;&lt;/p>

<p>you want your answer to be in seconds, so that means you have to reciprocate the speed of light and multiply it by the distance. However, you have to convert to the correct units. So here's how you set it up</p>

<p>[4.8 km][1000m/km][1/(3e8 m/s)][1000000000 ns/s]</p>

6

<p>thanks a lot. I ddidn't know that I was supposed to keep it in nanoseconds. I have ONE more question:</p>

<p>An automobile tire is rated to last for 55 000 miles. Estimate the number of revolutions the tire will make in its lifetime.</p>

<p>How can you do this if you dont know the circumference of the tire OR the speed? This is impossible</p>

7

<p>That problem cannot be done without using a value for the diameter/circumference. I think they want you to estimate a value for the circumference.</p>

8

<p>well what would I estimate it to? I am confused</p>

9

<p>well, that's why it's an estimate. you could google something, take a tape measure to your/your parents'/your neighbor's car...</p>

10

<p>what formula would i use?</p>

<p>mi/(height of tire) * 2 + 15?</p>

11

<p>Well, I won't give you a formula, but I'll give you enough info to figure it out.</p>

<p>First, 55,000 miles - thats the distance a point on the outside of the tire will move over its life. So, imagine you put a peice of gum on a tire...how far would it have travelled in one revolution? Simple - the circumfrence!.</p>

<p>So, if that wad of gum has to travel 55,000 miles, how many circumfrences - or revolutions - do you have to cover?</p>

<p>;)</p>

12

<p>Ok I got like 4300000 for my answer and I think that is a good guess.</p>

13

<p>Sounds reasonable. Just make sure you state your estimates/assumptions. :)</p>