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Sati [7]
3 years ago
15

Earth exerts a gravitational force on the moon. why doesn’t the force of gravity pull the moon into the earth?

Physics
1 answer:
Fittoniya [83]3 years ago
5 0
The moon is orbiting the Earth so the moon's acceleration is towards the Earth but its velocity is in tangent to the Earth. Therefore the moon is in a constant free fall around the Earth instead of getting pulled toward Earth it orbits around it.
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Are predators or their prey more likely to be successful?
mihalych1998 [28]
Predator because they are the one’s preying on the animals, the chances of the prey getting away is slim.
3 0
3 years ago
Read 2 more answers
A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 78 m horizontally
sertanlavr [38]

Answer:

Error

Explanation:

Speed= Distance / Time

Speed = 0.1

5 0
2 years ago
A coin of mass .005 kg dropped from a height of 3 m. How much potential energy does it have right before it hits the ground? Ass
k0ka [10]

Answer:

0 J

Explanation:

Gravitational potential energy is:

PE = mgh

where m is mass, g is acceleration due to gravity, and h is the height.

Right before the coin hits the ground, its height above the ground is 0 m, so the potential energy is:

PE = mgh

PE = (0.005 kg) (9.8 m/s²) (0 m)

PE = 0 J

6 0
3 years ago
A typical asteroid has an orbital period around the sun of 5.2 years. How far from the sun is this asteroid?
Elis [28]

Answer:

3 AU

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_a is the distance of the asteroid from the sun (orbital radius)

T_a=5.2 y is the orbital period of the asteroid

r_e = 1 AU is the orbital radius of the Earth

T_e=1 y is the orbital period the Earth

Solving the equation for r_a, we find

r_a = \sqrt[3]{\frac{r_e^3}{T_e^2}T_a^2} =\sqrt[3]{\frac{(1 AU)^3}{(1 y)^2}(5.2 y)^2}=3 AU

So, the distance of the asteroid from the Sun is exactly 3 times the distance between the Earth and the Sun.

3 0
3 years ago
What is the kinetic energy of a 4000kg bus traveling at 25m/s
bearhunter [10]
Kinetic Energy = 1/2mv^2
M= mass
V=velocity
4 0
4 years ago
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