Answer:
m = 1.99 kg = 2 kg
Explanation:
The moment of inertia of a bicycle rim about it's center is given by the following formula:

where,
I = Moment of Inertia of the Bicycle Rim = 0.21 kg.m²
r = Radius of the Bicycle Rim = Diameter of the Bicycle Rim/2
r = 0.65 m/2 = 0.325 m
m = Mass of the Bicycle Rim = ?
Therefore,

<u>m = 1.99 kg = 2 kg</u>
Answer:
5. Is greater than mg, always
Explanation:
If the cone has an inclination of angle β, the sum of forces will be:
x-axis (centripetal axis):
N*sin β = m*ax where ax is the centripetal acceleration
y-axis:
N*cos β - m*g = m*ay where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.
Solving for N:

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.
If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.
Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
Normal force is mass x gravity, so mass x 9.81
Answer:
a) 3673469.39 seconds
b) 6.61×10¹⁴ m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0.12×3×10⁸ m/s
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
Equation of motion

Time taken to reach 12% of light speed is 3673469.39 seconds

The distance it would have to travel is 6.61×10¹⁴ m