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NARA [144]
3 years ago
11

A 7.0kg skydiver is descending with a constant velocity

Physics
1 answer:
Vikentia [17]3 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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1 year ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
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A) 4.7 cm

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where

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a is the distance between the slits

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n = 5

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a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

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\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

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sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

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