Answer:

Explanation:
q = Charge of proton = 
r = Radius of circle = 
v = Velocity of proton = 
Magnetic moment is given by

The magnetic moment associated with this motion is 
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Answer:
625 W
Explanation:
Applying
P = W/t.................... Equation 1
Where p = power, W = Work, t = time
But,
W = Force (F) × distance (d)
W = Fd........................ Equation 2
Substitute equation 2 into equation 1
P = Fd/t.................... Equation 3
From the question,
Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds
Substitute these values into equation 3
P = (5000×30)/240
P = 625 Watt
Refer to the diagram shown below.
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N