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nika2105 [10]
2 years ago
14

A rock thrown with a velocity of 23.7 m/s horizontally off the top of a bridge falls 120 m to the water below. How far away does

the rock land?
Physics
1 answer:
Kipish [7]2 years ago
8 0

Answer:

d=117.2845834m

Explanation:

Let's start out with the vertical components

vi=0, vf=?, a=9.8, x=120, t=?

Let's find how long the rock was falling.

x=vi*t+.5*a*t^2

120=0+.5*9.8*t^2

120=4.9*t^2

t=4.9487166s

Since we know the horizontal velocity is constant, we use the formula v=d/t

23.7=d/4.9487166

d=117.2845834m

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T = \frac{ Id^{2}\theta }{dt^{2} }

-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }

\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )

By Comparing the above equation with the SHM equation

\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta

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Frequency = \frac{\mu}{2\pi }

=\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}

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Find the polar coordinates corresponding to a point located at (−6.48, 12.58) in Cartesian coordinates
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Explanation:

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