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liberstina [14]
3 years ago
9

A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E

arth to the surface. If the satellite had a weight at the surface of 4000 N, what is the force of gravity (weight) of the satellite when it is in its orbit? Give your answer in newtons, N.
Physics
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

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A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and
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Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

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or

θ=ωt

and we know

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From this we have

angular velocity ω = 1.4 revolutions per sec =  1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r  in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

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