A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

which can be rewritten as

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

So, we get:

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B) 
The frequency of revolution of a particle in uniform circular motion is

where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:

<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here
then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))
The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>
9.3 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 70kg
Mass 2 = 2000kg
distance = 1m
Unknown:
force between them =
Solution:
The force between the two masses will be a gravitational force of attraction.
F = 
G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²
r is the distance between the two masses
Substituting the parameters:
F =
= 9.3 x 10⁻⁶N
Learn more:
Universal gravitation constant
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