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allsm [11]
3 years ago
9

A 0.10 kg ball of dough is thrown straight up into the air with an initial speed of 15 m/s.

Physics
1 answer:
Reptile [31]3 years ago
6 0
<span>Mass of the ball is m = 0.10kg Initial speed of the Ball v = 15m/s a. When the ball is at maximum height the velocity is 0 Momentum of ball = mass x velocity Momentum = 0.10kg x 0 = 0 b. Getting the maximum height, Using the conservation of energy equation KEinitial = mgh 1/2mVin^2 = mgh => h = v^2/2g h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m Applying the conservation of energy equation at halfway V^2 = 2gh V = square root of (2x9.8x5.96) => V = square root of (116.816) So the velocity at the half way V = 10.81 m/s Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
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Answer:

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Explanation:

Given

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Step two:

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Explanation:

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m1= 900kg

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Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

r= \sqrt{\frac{Gm1m2}{F} }

r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m

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