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mixer [17]
3 years ago
9

A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to

the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?

Physics
1 answer:
rodikova [14]3 years ago
8 0

Answer:F=40.09 N

Explanation:

Given

weight of crate W=50 N

Inclination \theta =37^{\circ}

Frictional Force f=10 N

as the crate is moving with constant velocity therefore net Force on crate is zero

F-50\sin (37)-f=0

F=50\sin (37)+10

F=30.09+10

F=40.09 N

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The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

F=mg (1)

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m is the mass of the astronaut

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F=\frac{GMm}{r^2}

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F=\frac{GMm}{R^2}=640 N

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

r'=3R+R=4R

Therefore, the new weight is

F'=\frac{GMm}{(4R)^2}=\frac{1}{16}\frac{GMm}{R^2}=\frac{F}{16}

Which means that his weight has decreased by a factor 16: therefore, the new weight is

F'=\frac{640}{16}=40 N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

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