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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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Darya [45]

Answer:

Explanation:

If Ek is the kinetic energy and m is the mass and v is the velocity then v can be calculated as follows

Ek= 1/2 ×( m × v² )

2Ek= mv²

2Ek/m = v²

v =√(2Ek/m)

m = 0.1 kg

v= √(2x8/0.1)= 12.65 m/s

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3 years ago
What factors affect attractive force
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Two Factors That Affect How Much Gravity Is on an Object. Gravity is the force that gives weight to objects and causes them to fall to the ground when dropped. Two major factors, mass and distance, affect the strength of gravitational force on an object.

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3 years ago
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Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

5 0
3 years ago
The ground state energy of an electron in a one-dimensional trap with zero potential energy in the interior and infinite potenti
ElenaW [278]

Answer:

0.5 eV

Explanation:

E_1 = Initial potential energy = 2\ eV

E_2 = Final potential energy

L_1 = Initial width

L_2 = Final width = 2L_1

Energy of an electron in a one-dimensional trap is given by

E=\dfrac{n^2h^2}{8mL^2}

From the equation we get

E\propto \dfrac{1}{L^2}

So,

\dfrac{E_1}{E_2}=\dfrac{L_2^2}{L_1^2}\\\Rightarrow E_2=\dfrac{E_1L_1^2}{L_2^2}\\\Rightarrow E_2=\dfrac{2L_1^2}{4L_1^2}\\\Rightarrow E_2=0.5\ eV

The ground state energy will be 0.5 eV

6 0
3 years ago
A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi
Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

y=V_{o}sin\theta t-\frac{gt^{2}}{2}   (1)

V_{y}=V_{oy}-gt   (2)

Where:

y  is the height of the cat  

V_{oy}=V_{o}sin\theta is the cat's initial velocity

\theta=60\°

g=9.8m/s^{2}  is the acceleration due gravity

t is the time

V_{y} is the y-component of the velocity

Now the cat will have its maximum height y_{max} when V_{y}=0. So equation (2) is rewritten as:

0=V_{oy}-gt   (3)

Finding t:

t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}   (4)

t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}   (5)

t=0.24 s   (6)

Substituting (6) in (1):

y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}   (7)

Finally:

y_{max}=0.287 m \approx 0.29 m   (8)

3 0
3 years ago
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