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3 years ago

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An autographed baseball rolls off of a 0.93 m high desk and strikes the floor 0.56 m away from the desk. How fast was it rolling

on the desk before it fell? The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.

1 answer:

Dafna1 [17]3 years ago

6 0

First of all, how long does it take it to fall 0.93m to the floor ?

Distance falling from rest = (1/2) · (g) · (T²) <== memorize this formula

(0.93 m) = (1/2) (9.8m/s²) (T²)

Divide each side by (4.9 m/s²) : T² = (0.93/4.9) = 0.1898 s²

Square root each side: T to hit the floor = 0.436 second

It got 0.56m away from the desk horizontally in 0.436 sec.

Speed = (distance) / (time) <=== memorize this formula

Speed = (0.56 m) / (0.436 sec)

<em>Speed = 1.28 m/s</em>

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Find enclosed file with a drawing and the calculations of all angles and all distances.

These are the results that you will find in it:

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3 years ago

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Answer:

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$V$ - Volume, measured in cubic meters.

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta h$ - Vertical displacement, measured in meters.

If we know that $n = 100$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 0.02\,m^{3}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 1\,m$ , then the work done is:

$\Delta W = (100)\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot (0.02\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1\,m)$

$\Delta W = 19614\,J$

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