Answer:
Explanation:
Let g = 10m/s2. Gravity acting on the rod:
G = mg = 10M
Suppose the rod is uniform, then we can treat gravitational torque as gravity force acting on the center of mass, which is rod's midpoint
T = 0.5GR = 0.5*10M*4 = 20M
The moment of inertia of the rod about 1 end is
Then the angular acceleration as soon as it releases is
Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
Voltage = current x resistance
voltage = 24 x 7 = 168 volts
Explanation:
The expression for the magnetic field around the current carrying wire is given by :
Here,
I is electric current and r is distance from the wire.
The factors affect the strength of magnetic field around current carrying wire are :
- Electric current
- Distance from wire
