The question is incomplete.
The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.
Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.
Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.
Answer: The Electrostatic force of attraction or repulsion between two charges shows that the Newton's third law applies to electrostatic forces.
Explanation: Consider two Oppositely charged charges separated by distance d.
The electrostatic force exerted by charge 1 on charge 2 is.
By Coulomb's Law :
F1 = k
.....................................(1)
The electrostatic force exerted by charge 2 on charge 1 is.
F2 = - k
................................. (2)
negative sign shows that force are in opposite direction.
From Equation 1 and 2
F1 = - F2
Which implies Newton Third law.
So for a minute lets ignore the 880 km/h. If it took 4 hours and she flew at 600 km/h 600*4=2400. Now lets Look at the 880 bit. If it took 4 hours and she where to fly at 880 it would've been 880*4=3520. Lets do 2400-600=1800, now we've got the 600 kmh bit done. Now lets see if you fly 880 km/h for one hour then you add 1800+880=2680.
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Answer:
vo=5.87m/s
Explanation:
Hello! In this problem we have a uniformly varied rectilinear movement.
Taking into account the data:
α =69.2
vf = 10m / s
h=2.7m
g=9.8m/s2
We know we want to know the speed on the y axis.
We calculate vfy
vfy = 10m / s * (sen69.2) = 9.35m / s
We can use the following equation.

We clear the vo (initial speed)


vo=5.87m/s
Answer:
This reduces the average force applied during the landing process/ or you can say it reduces the impact your body takes.
Explanation: