Answer:
option C
Explanation:
given,
diameter of circular room = 8 m
rotational velocity of the rider = 45 rev/min
=
=4.712 rad/s
here in this case normal force is equal to centripetal force
N = m r ω²
N = m x 4 x 4.712²
N = 88.83m
frictional force = μ N
= 88.83m x μ
now, for the body to not to slide
gravity force is equal to frictional force
m g = 88.83 m x μ
g = 88.83 x μ
9.8 = 88.83 x μ
μ = 0.11
hence, the correct answer is option C
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = --------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = -------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v =
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s