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2 years ago

A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body

Physics

1 answer:

Aleksandr-060686 [28]2 years ago

4 0

P=30 kg•m/s. This will be your correct answer.

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

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Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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2 years ago

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Answer:

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

Explanation:

Gravitational force between two objects of masses $m_{1}, m_{2}$ kept at a distance r is given by the formula

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$

Here , $m_{1}$ = 2m

$m_{2}$ = $\frac{m}{2}$

Thus , F = $\frac{-G.2m.\frac{m}{2} }{r^{2} }$

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

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A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body​

A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body