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just olya [345]
3 years ago
12

When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. If you do twice as much work w

hen throwing the ball, does it go twice as fast? Explain. Yes. Twice as much work will give the ball twice as much kinetic energy. Since KE is proportional to the speed, the speed will double as well. Yes. Twice as much work will give the ball four times as much kinetic energy. Since KE is proportional to the speed squared, the speed will be the square root of 4, or twice as fast. No. Twice as much work will give the ball four times as much kinetic energy. Since KE is proportional to the speed, the speed will be four times larger. No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be 2 times larger.
Physics
1 answer:
aniked [119]3 years ago
5 0

Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be sqrt{2} times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

W=K

So when the work done doubles, the kinetic energy doubles as well:

2W = 2 K

However, the kinetic energy is given by

K=\frac{1}{2}mv^2

where

m is the mass of the ball

v is its speed

We see that the kinetic energy is proportional to the square of the speed, v^2. We can rewrite the last equation as

v=\sqrt{\frac{2K}{m}}

which also means

v=\sqrt{\frac{2W}{m}}

If the work is doubled,

W'=2W

So the new speed is

v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v

So, the speed is \sqrt{2} times larger.

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What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

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Therefore, the magnification of an astronomical telescope is -30.83.

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3 years ago
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2 years ago
What's the equivalent resistance to the resistances shown?<br><br>A) 5<br>B) 6<br>C) 2<br>D) 3
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