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densk [106]
3 years ago
7

What is the time required for an object starting

Physics
1 answer:
PtichkaEL [24]3 years ago
8 0
(3) 10.1 second Using equation of motion 500 = (0.5)(9.81)t^2. Rearranging, t = sqrt(1000/9.81) = 10.1s
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A 3.1-kilogram gun initially at rest is free to
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The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
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6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory
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Answer:

Engular velocity: w=1,24*10^{-3}/s

Linear velocity: V=7905 m/s

The time it takes:

t=5060s=84min

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The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1)a=r*w^{2}

Solving for w:

(2)w=\sqrt{\frac{a}{r} }

Replacing a=9,8m/s2 and r=6,375,000m:

(3)w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s

And the angular velocity relates to the linear velocity:

V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s

The perimeter of the orbit is:

P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m

The time it takes:

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