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Sliva [168]
3 years ago
13

The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is transported f

rom the moon to the earth, which properties of the rock change? The gravitational field on the surface of the earth is stronger than that on the surface of the moon. If a rock is transported from the moon to the earth, which properties of the rock change? mass only weight only both mass and weight neither mass nor weight
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer: Weight only.

Explanation: Mass is a measure of the amount of matter in an object. Weight is a measure of the gravitational force exerted on the material in a gravitational field. Mass and weight are proportional to each other, with the acceleration due to gravity as the proportionality constant.

If a rock is transported from the moon to the earth, the mass is constant for the object but the weight will depends on the locations of the object. The gravitational acceleration would change because the radius and mass of the Moon is different from the Earth.

Thus, the object (rock) has <em>mass, m</em> both on the surface of the Earth and the surface of the Moon; but it will <em>weight</em> much less on the surface of the Moon as the Moon's surface gravity is 1/6 of the Earth.

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Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o
Sholpan [36]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = f_o\times\frac{V+v_o}{ V-v_s}

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  f\times \frac{340+6}{340-10}

= f\times \frac{346}{330}

So m = 346 , n = 330 .

8 0
3 years ago
A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide
-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0
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Answer:

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