Here if we assume that there is no air friction on both balls then we can say
now the acceleration is given as
so here both the balls will have same acceleration irrespective of size and mass
so we can say that to find out the time of fall of ball we can use
now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size
Answer:
i. The radius 'r' of the electron's path is 4.23 × m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r =
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ =
(since it enters perpendicularly to the field), q = e = 1.6 ×
C and m = 9.11 ×
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 ×
× 1.63 ×
)
= 1.10231 × ÷ 2.608 ×
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f =
= (1.6 × × 1.63 ×
) ÷ (2 ×
× 9.11 ×
)
= 2.608 × ÷ 5.7263 ×
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.