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2 years ago

15

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

1 answer:

lisabon 2012 [21]2 years ago

6 0

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

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Answer:

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Explanation:

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substituting the values for $m$ and $d$ :

$\theta=sen^{-1}(\frac{(1)\lambda}{(1.33x10^{-6}m)})$

And we need to find two angle values: one for when the wavelength is 400nm and one for when it is 700 nm. So we will get the angle range

$\theta=sen^{-1}(\frac{(400x10^{-9})}{(1.33x10^{-6}m)})=17.50$

and

$\theta=sen^{-1}(\frac{(700x10^{-9})}{(1.33x10^{-6}m)})=31.76$

The range of angles is from 17.50° to 31.76°

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