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3 years ago

12

What is the acceleration of a baseball that has a final velocity of 15.0 m/s when it crosses the plate 2.0 seconds after leaving

the pitchers hand? The initial velocity is 0.0 m/s and the baseball is traveling forward.

1 answer:

Leokris [45]3 years ago

3 0

I think it’s A pretty sure...

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The gravitational potential is a distance z=0.17 m away from a distribution of mass that has the following equation:

sveta [45]

gravitational potential is given as

$V = \frac{GM}{R^2}(R^2 + Z^2)^0.5$

$E = -\frac{dV}{dz}$

$E = -\frac{GM}{R^2}\frac{d}{dz}(R^2+z^2)^0.5$

$E = -\frac{GM}{R^2}*0.5(R^2+z^2)^-0.5*2z$

$E = -\frac{GM*z}{R^2*(R^2 + z^2)^0.5}$

now plug in all values given

M = 110 kg

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z = 0.17 m

$E = -\frac{6.67 * 10^{-11}* 110*0.17}{0.55^2*(0.55^2 + 0.17^2)^0.5}$

$E = 7.16 * 10^{-9} N/kg$

so above is the field intensity

4 0

3 years ago