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atroni [7]
3 years ago
6

A charged rod is brought near one end of a long, uncharged metal block. Students want to experimentally measure the resulting ch

arge distribution along the entire length of the block. They have a small, positively charged sphere on a string that can be used as a test charge with negligible effect on the other charges. They will observe whether the sphere is attracted or repelled when held near the rod. Which of the following describes and justifies a procedure that will provide data to determine the entire charge distribution?
a. Hold the sphere near the end of the block closest to the rod, because that will give experimental data about both ends of the block.
b. Hold the sphere near each end of the block, because that will give experimental data about both ends of the block.
с. Hold the sphere near each end of the block and near the block's middle, because that will give experimental data about the area along the length of the block.
d. Hold the sphere near each end of the block and at a number of points along the length of the block, because that will give experimental data for the whole block.
Physics
1 answer:
quester [9]3 years ago
7 0

Answer:

d

Explanation:

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Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

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If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

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Replacing into the first equation:

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Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

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Answer:....

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