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4 years ago

A 7.0kg skydiver is descending with a constant velocity

Physics

1 answer:

Vikentia [17]4 years ago

6 0

Answer:

The air resistance on the skydiver is 68.6 N

Explanation:

When the skydiver is falling down, there are two forces acting on him:

- The force of gravity, of magnitude $mg$ , in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)

- The air resistance, $R$ , in the upward direction

So the net force on the skydiver is:

$F=mg-R$

where

m = 7.0 kg is the mass

$g=9.8 m/s^2$

According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):

$F=ma$

In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:

$a=0$

Therefore the net force is zero:

$F=0$

And so, we have:

$mg-R=0$

And so we can find the magnitude of the air resistance, which is equal to the force of gravity:

$R=mg=(7.0)(9.8)=68.6 N$

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Answer:

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3 years ago

A solid disk of radius 5.50 cm and mass 1.25 kg , which is rolling at a speed of 1.50 m/s , begins rolling without slipping up a

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Answer:

Explanation:

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3 years ago

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Answer:

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Explanation:

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$F_{c}=ma_{c}$

$F_{c}=m\frac{v^{2}}{R}$

Where:

R is the length of the rope

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v is the speed of the object

If we want to reduce the centripetal force in half, we can duplicate the length of the rope (R'=2R), which means:

$F'_{c}=m\frac{v^{2}}{R'}$

$F'_{c}=m\frac{v^{2}}{2R}$

$F'_{c}=\frac{1}{2}F_{c}$

I hope it helps you!

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3 years ago

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4 years ago