Explanation:
Mass of the wheel, m = 49 kg
Radius of the hoop, r = 0.73 m
Initial angular speed of the wheel, 
Final angular speed of the wheel, 
Time, t = 22 s
(a) If I is the moment of inertia of the hoop. It is equal to,



We know that the work done is equal to change in kinetic energy.



W = -1858.05 Joules
(b) Let P is the average power. It is given by :


P =84.45 watts
Hence, this is the required solution.
Answer:
7.78 * 10³ m/s
Explanation:
Orbital velocity is given as:
v = √(GM/R)
G = 6.67 * 10^(-11) Nm/kg²
M = 5.98 * 10^(24) kg
R = radius of earth + distance of the satellite from the surface of the earth
R = 2.15 * 10^(5) + 6.38 * 10^(6)
R = 6.595 * 10^(6) m
v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))
v = √(6.048 * 10^7)
v = 7.78 * 10³ m/s
Answer:
Explanation:
The magnitude of the electric force on this charged particle A depends upon the following
5. the distance between the point charge Q and the charged particle A
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A