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:
Explanation:
We could use the following suvat equation:
where
s is the vertical displacement of the coin
v is its final velocity, when it hits the water
t is the time
g is the acceleration of gravity
Taking upward as positive direction, in this problem we have:
s = -1.2 m
And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:
where the negative sign means the direction is downward.
Water expands when it freezes (that's why you should never put closed, fully filled water bottles in the freezer !)
<span>The average weather of a particular place is "Climate"
In short, Your Answer would be Option B
Hope this helps!</span>
Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = 
m
we substitute
q = 
m
q = 
m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one
