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VMariaS [17]
1 year ago
10

The moon has a mass of 1×10^22 kg, and the gravitational field strength at a distance r from the planet is 0.001 n/kg. what is t

he gravitational force exerted on the moon while it is in orbit around the planet?
Physics
1 answer:
CaHeK987 [17]1 year ago
4 0

When I divided by 6, I got W(moon) = mg(moon) = 10 x 10/6 = 16.7N (the moon's gravity is only 1/6 that of Earth in this case). 100N is equal to W(Earth) = mg(Earth) = 10 x 10. On the moon and the Earth, m = 10 kg.

How can you determine the strength of Earth's gravitational field at the moon?

M = 6 10 24 K g is the mass of the earth, and g = - G M R 2 is the gravitational field of the earth at the point of the moon. The distance between the earth and moon is 84 10 8 meters. field of gravitation.

How do you determine the distance between Earth and the moon where the gravitational field is at its weakest?

You only discover points where the forces from all of the different local bodies are in balance, or equal to each other, because there is no place in the universe where the gravitational field intensity is zero. All bodies, not just the Earth and Moon, must be taken into account in the computation to arrive at these positions.

To know more about gravitational field strength at moon visit;

brainly.com/question/20909424

#SPJ4

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Name few biodegradable materials that get decomposed in a week​
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Answer:

Cardboard, Paper towels, Food waste, and wooden based items

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3 years ago
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If the coefficient of static friction is 0.40, and the same ladder makes a 51° angle with respect to the horizontal, how far alo
inysia [295]

To solve this problem, we apply the concepts related to the sum of forces and balance in a diagram that will be attached, in order to identify the behavior, direction and sense of the forces. The objective is to find an expression that is in terms of the mass, the angle, the coefficient of friction and the length that allows us to identify when the ladder begins to slip. For equilibrium of the ladder we have,

\sum F_x = 0

\sum F_y = 0

\sum M_o = 0

Now we have that

f_1 = N_2

N_1 = mg

And for equilibrium of the two forces we have finally

mgdcos\theta = N_2lsin\theta

Rearranging to find the distance,

d = \frac{N_2}{mg}ltan\theta

d = \frac{f_1}{mg}ltan\theta

So if we have that the frictional force is equivalent to

f_1 = \mu N_1

f_1 = \mu mg

f_1 = (0.4)(57*9.8)

f_1 = 223.44N

With this value we have that

d = \frac{(0.4)(57)(9.8)}{57*9.8}(7.5) tan(60\°)

d = 5.19m

Therefore can go around to 5.19m before the ladder begins to slip.

5 0
3 years ago
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

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The sound waves from a noisy jet travel from the air into water. Which property of the wave will not change?​
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The frequency of the wave will not change. Since the change in medium doesn't affect the source of the waves, the frequency of those waves do not change.

Hope this helps! :)

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