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allochka39001 [22]
3 years ago
5

A rocket experiences a constant force even as the amount of fuel in its fuel tanks decreases. What happens to the acceleration o

f the rocket as it runs out of fuel?
A - The acceleration decreases because the mass increases.B - The acceleration increases because the mass increases.C - The acceleration decreases because the mass decreases.D - The acceleration increases because the mass decreases.
Physics
2 answers:
prohojiy [21]3 years ago
7 0

We can answer this question by looking at Newton's second law:

F=ma

which can be rewritten as

a=\frac{F}{m}

where F is the force experienced by the rocket, m is its mass, a is its acceleration. In the rocket's case, the mass of the rocket decreases (because the fuel in the tank decreases), while the force remains constant, so the ratio F/m increases, and therefore the acceleration of the rocket increases.

Therefore, the correct answer is

D - The acceleration increases because the mass decreases.

Elanso [62]3 years ago
3 0
The correct answer is D
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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.

Downward motion:

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

Since, the motion is downward, final velocity must be negative. So,

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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3 years ago
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