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Afina-wow [57]
3 years ago
14

hen a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant

force acts upon an object with mass , the acceleration of the object is . If the same force acts upon another object whose mass is , what is this object's acceleration?
Physics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer:

Incomplete question

Complete question:

a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an object with mass 44kg , the acceleration of the object is 4m/s². If the same force acts upon another object whose mass is 11kg what is this object's acceleration?

Answer: 8m/s²

Explanation:

From the statement we deduced that acceleration varies inversely with mass where force was kept constant.

Therefore,

F/m = a or F = ma

For the first statement, substituting the mass and acceleration gives:

F = 44 x 4 = 88N

Applying the force above to the second mass gives us:

a = 88/11 = 8m/s²

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A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
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Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

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Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

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v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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