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Gnesinka [82]
3 years ago
13

A ball of mass 0.05 kg strikes a smooth wall normally four times in 2 second with a velocity of 10m/s. Each time the ball reboun

ds with the same speeds of 10m/s. Calculate the average force on the wall
Physics
1 answer:
asambeis [7]3 years ago
3 0

Answer" The average force on the wall be equal to 0.5 N.

Explanation:

To calculate the average force on the wall, we use the following equation:

Ft=m\Delta v\\\\F=\frac{m(v_2-v_1)}{t}

where,

F = average force = ?N

m = mass of the ball = 0.05kg

v_1 = Initial velocity of the ball = -10m/s (negative sign because the ball rebounds)

v_2 = Final velocity of the ball = 10m/s

t = time taken by the ball = 2s

Putting values in above equation, we get:

F=\frac{0.05kg[10-(-10)]m/s}{2s}\\\\F=\frac{0.05kg(20m/s)}{2s}\\\\F=0.5kgm/s^2=0.5N

Hence, the average force on the wall will be equal to 0.5N.

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Distance fallen = 1/2 ( V initial + V final ) *t
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To find distance fallen, we need to find V final
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This way plugging into distance equation is actually the long way. A faster way is to plug the values into
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We won't need to find V final using another equation.

But anyways, good luck!



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