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.
The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this
Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²
Height = (4.9 m/s²) (5.76 s²)
Height = (4.9/5.76) meters
Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)
Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.
Answer:
.067 so C
Explanation:
I asked my sister who is in 2nd grade and she said it was right so you are good! =). have a great day!
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:

, where

is the coefficient of friction and

is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:

And so, this is the force that the worker must apply to the crate.