Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²
B is the answer, I’m really good at this subject
The frictional force is in the opposite direction
Answer:
Explanation:
Acceleration of particle A is 7.3 times the acceleration of particle B.
Let the acceleration of particle B is a, then the acceleration of particle A is
7.3 a.
Let the period of particle A is T and the period of particle B is 2.5 T.
Let the radius of particle A is RA and the radius of particle B is RB.
Use the formula for the centripetal force

So, 
The ratio of radius of A to the radius of B is given by


RA : RB = 1.17