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allsm [11]
3 years ago
5

A 125 kg linebacker runs to the right at 2.5 m/s and an 80 kg quarterback runs to the left at 3 m/s. They collide in midair and

the linebacker holds on to the quarterback so they move off together. What is their velocity (magnitude and direction) immediately after this collision
Physics
1 answer:
Alexxx [7]3 years ago
6 0

Answer: 0.35 m/s to the right.

Explanation:

Let's define M1 and V1 as the mass and velocity of the linebacker, and define the right as the positive side.

M1 = 125kg

V1 = 2.5m/s

M2 and V2 as the velocity and mass of the quarterback.

M2 = 80kg

V2 = -3m/s

where V2 has a minus sign because the quarterback is running to the left.

We know that after the collision, both players keep stick togheter, so we can think this as an perfectly inleastic collision, where the velocity of both players can be obtained by:

v = (M1*V1 + M2*V2)/(M1 + M2) = (125kg*2.5m/s - 80kg*3m/s)/(125kg + 80kg)

v = (72.5/205) m/s = 0.35m/s

Because this velocity is positive, this means that both players are moving to the right.

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Answer:

Hydrogen and helium compounds.

Explanation:

We know that the solar System was formed around <u>4.6 billion years ago, </u>due to the gravitational collapse of a giant interstellar molecular cloud.

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I hope it helps you!

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A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.
Sunny_sXe [5.5K]

Answer: 240\ rad/s^2

Explanation:

Given

Length of beam l=2\ m

mass of beam m=5\ kg

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

\tau =F\times l=200\times 2=400\ N.m

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2

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