I believe this is what you have to do:
The force between a mass M and a point mass m is represented by

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ =
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ =
F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
Answer:
v₀ = 16.55 m/s
Explanation:
This motion of the ball can be modeled as a projectile motion with following data:
R = Range of Projectile = 27.5 m
θ = Launch Angle = 50°
g = acceleration due to gravity = 9.81 m/s²
v₀ = Initial Speed of Ball = ?
Therefore, using formula for range of projectile, we have:

<u>v₀ = 16.55 m/s</u>
Answer:
40 km/h
Explanation:
First...
Look at the formula speed is equal to the distance over time or s = d/t.
Next...
Use the formula: 240/6.0
Finally...
Solve: 40
So the answer: 40 km/h