<u><em>PRIMARY </em></u>Waves Are Detected First Because They Move So Fast.
<u><em>RIGHT</em></u> Angles To The Direction of Movement.
A Kind Of Scale Used To Measure The Amount of Seismic Energy Released By An Earthquake <u><em>RICHTER SCALE</em></u>
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:
it is a force
Explanation:
a force is a push or a pull
Thank you for posting your Physics question here. I hope the answer helps. Upon calculating the ramp with the horizontal the answer is 20.49 Deg. Below is the solution:
Y = 7 m.
<span>r = 20 m. </span>
<span>sinA = Y/r = 7/20 = 0.35. </span>
<span>A = 20.49 Deg.</span>
Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
