A)planets
b)the sun
c)moons
e)comets
f)asteroids
Answer:
Did you ever get the answer?
Explanation:
Answer:
No the gravity of the moon pulls the water making high tide
Explanation:
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:

Answer:
PE=0.92414J and KE=0.28175J
Explanation:
Gravitational potential energy=mass*gravity*height
PE=mgh
Data,
M=0.046kg
H=2.05m
g=9.8m/s^2
PE=0.046kg * 9.8m/s^2 * 2.05m
PE =0.92414J
KE=1/2mv^2
M=0.046kg
V=3.5m/s
KE=[(0.046kg)*(3.5m/s)^2]\2
KE=0.28175J