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Step2247 [10]
2 years ago
9

Determine the orbits period of the moon when the distance between the earth and the moon is 3.82 x 10 to the power of 8

Physics
1 answer:
alexandr402 [8]2 years ago
4 0

Answer

From

V=Distance/time

The distance round a circular path or Object is 2πr

so

v=2πr/T

Making T(period) subject

T=2πr/v

where v is the linear velocity.

We don't have the velocity but we can get it.

The Moon and Earth exert gravitational force on each other and the Moon is kept in Orbit by Centripetal force.

Since the Moon doesn't fly out of Orbit... it must mean that the Gravitational force being exerted on it by the Earth is equal to its centripetal Force.

Equating Both (Fg=Fc)

Fg=GMm/R²

Fc=mv²/r

GMm/R² = mv²/R

Canceling out "m" and "r" on both sides

we're left with

V²=GM/R

V=√GM/R

Where M= Mass of Earth (5.98x10^24)

R=Distance between the center of earth and the Moon

G= Gravitational Constant(Value 6.67x10^-11)

V=√6.67x10^-11 x 5.98x10^24/(3.82x10^8)

V=1021.84meters per second.

Now

T=2πr/v

=2π x 3.82x10^8 / 1021.84

T=2.35x10^6seconds

Converting to days by dividing by(24 x 3600)

You have

T=27.2days approx.

✅✅✅

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But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

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Answer:

a)Distance traveled during the first second = 4.905 m.

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c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

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a) We have equation of motion

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     Here u = 0, and a = g

              S = 0.5gt²

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              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

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      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

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      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

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   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
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