Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Answer:
4.25 m/s
Explanation:
Force, F = 22 N
Time, t = 0.029 s
mass, m = 0.15 kg
initial velocity of the cue ball, u = 0
Let v be the final velocity of the cue ball.
Use newton's second law
Force = rate of change on momentum
F = m (v - u) / t
22 = 0.15 ( v - 0) / 0.029
v = 4.25 m/s
Thus, the velocity of cue ball after being struck is 4.25 m/s.
Answer:
44(speed) x 10(time) = 440(distance). So the answer is 440 meters.
