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3 years ago

Q1) An average force of 50.0 N is exerted on a 4.0-kg cart for 2.0 seconds.

Physics

1 answer:

lozanna [386]3 years ago

3 0

<h3>a. The impulse</h3>

The impulse is 100.0 Ns

The impulse I = Ft where

F =average force = 50.0 N and
t = time = 2.0 s

Substituting these values into the equation, we have

I = Ft

I = 50.0 N × 2.0 s

I = 100.0 Ns

The impulse is 100.0 Ns

<h3>b. Change in momentum</h3>

The change in momentum is 100 kgm/s

Since change in momentum Δp = I where I = impulse.

Since I = 100.0 Ns,

Substituting this into the equation, we have

Δp = I

= 100.0 Ns

= 100 kgm/s

The change in momentum is 100 kgm/s

<h3>c. Mass's change in velocity</h3>

The change in velocity is 25.0 m/s

Since change in momentum Δp = mΔv where

m = mass = 4.0 kg and
Δv = change in velocity.

Making Δv subject of the formula, we have

Δv = Δp/m

Substituting the values of the variables into the equation, we have

Δv = Δp/m

Δv = 100.0 kgm/s/4.0 kg

Δv = 25.0 m/s

The change in velocity is 25.0 m/s

Learn more about impulse here:

brainly.com/question/25700778

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Answer:

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Explanation:

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q_A = 3/8 Q

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Hey there

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