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3 years ago

14

Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially

elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision:______
a. each car has half the momentum.
b. car A stops and car B has momentum m v.
c. car A stops and car B has momentum 3m v.
d. the momentum of car B is three times as great in magnitude as that of car A.
e. each car has half of the kinetic energy.

1 answer:

mixer [17]3 years ago

3 0

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

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State whether the following statement are true or false .

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1. It’s true
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3 years ago

It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit

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Answer:

$a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

Explanation:

According to Newton's Third Equation of motion

$V_f^2-V_i^2=2as$

Where:

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$V_i$ is the initial velocity

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s is the distance

In our case:

$V_f=V_{escape}, V_i=0,s=4 m$

So Equation will become:

$V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

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3 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

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The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

$a_x$ is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

Learn more about force of friction:

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3 years ago

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2 years ago

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Explanation:

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3 years ago