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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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3 0
3 years ago
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A boat can travel 2.30 m/s in still water. If the boat points its prow directly across a stream whose current is 1.80 m/s , what
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Answer:

A boat can travel 2.30 m/s in still water. (a) If the boat points its prow directly across a stream whose current is. ... (a) What is the velocity (magnitude and direction) of the boat relative to the shore? ... The boat's velocity with respect to the shore, , is the sum of its velocity with respect to the water

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3 years ago
Define real machine.​
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Answer:

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Explanation:

Have A Wonderful Day !!

3 0
3 years ago
Every summer I drive to Pennsylvania. It is 895 km to get there. If I average 100 km/hr, how much time will I spend driving?
Ivan

Answer the time you bee spending driving iss 795 because 895-795=100

Explanation:

6 0
2 years ago
The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo
a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s

The tire angular velocity before stopping is:

\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s

Also its angular decceleration:

\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2

Using the following equation motion we can findout the angle it makes during the deceleration:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

-114^2 = 2*(-14.75)*\Delta \theta

\Delta \theta = 440rad or 440/2π = 70 revelutions

4 0
3 years ago
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