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3 years ago

When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart

Physics

2 answers:

Karolina [17]3 years ago

7 0

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

$I=\Delta p=m\Delta v$

where

m = 5.0 kg is the mass of the cart

$\Delta v=2.2 m/s$ is its change in speed

Substituting numbers into the equation, we find

$I=(5.0kg)(2.2 m/s)=11 kg m/s$

ivolga24 [154]3 years ago

7 0

Solution:

In this question we have given

mass of Cart=5Kg

change in speed, $\Delta v=2.2\frac{m}{s}$

We have to find, Impulse, I=?

Impulse :

Impulse of a body is equla to its change in momentum.

Therefore,

Impulse of cart is equal to its change in momentum:

$I=\Delta p\\I=m\Delta v$ ................(1)

Put values of $\Delta v$ , and m in eq(1)

$I=(5.0kg)(2.2\frac{m}{s} )\\I=11 kg\frac{m}{s}$

Therefore, the impulse of the cart, $I=11 kg\frac{m}{s}$

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