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dexar [7]
3 years ago
6

A planet has a gravitational acceleration on its surface of 2.2 times Earth's gravitational acceleration on its surface. The pla

net's radius is five times Earth's radius. What is the mass of the planet, in terms of Earth masses, ME?
Physics
1 answer:
lesantik [10]3 years ago
5 0

Answer:

The mass of the planet is 55 times the mass of earth.

Explanation:

From the inverse-square gravitation law,

F = (GMm/r²)

If the weight of a body (the force with which the earth attracts a body to its centre) is to be calculated,

F = mg

m = mass of the body,

g = acceleration due to gravity

mg = (GMm/r²)

G = Gravitational constant

M = mass of the earth

m = mass of body

r = distance between the body and the centre of the earth = radius of the earth

The acceleration due to gravity is given by

g = (GM/r²)

Making the mass of the earth, the subject of formula

M = (gr²/G) (eqn 1)

So, the planet described,

Let the acceleration due to gravity on the planet be g₁

Mass of the planet be M₁

Radius of the planet be r₁

g₁ = 2.2g

r₁ = 5r

M₁ = ?

Note that the gravitational constant is the same for both planets.

So, we can write a similar expression for the planet's acceleration due to gravity

g₁ = (GM₁/r₁²)

Substituting all the parameters known in terms of their corresponding earth values

2.2g = [GM₁/(5r)²]

2.2g = [GM₁/25r²]

M₁ = (55gr²/G)

Recall the expression for the mass of the earth

M = (gr²/G)

M₁ = 55 M

The mass of the planet, in terms of Earth masses = 55M

The mass of the planet is 55 times the planet of earth.

Hope this Helps!!!

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A) 4.7 cm

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In this problem,

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

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B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

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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}

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a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

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I=\Delta p=p_{2}-p_{1} (1)

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I is the impulse

\Delta p is the change in momentum

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p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

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I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

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