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olya-2409 [2.1K]
2 years ago
14

An object with a mass of 70 kilograms is supported at a height 8 meters above the ground. What's the potential energy of the obj

ect with respect to the ground? A. 3,430 J B. 2,058 J C. 5,488 J D. 560 J
Physics
1 answer:
alexira [117]2 years ago
5 0
Ep= mgh

70 x 9.8 x 8

Ep= 5,488J
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Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

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V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

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\tau =  2.1161 s

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<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

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