Answer:
0.006075Joules
Explanation:
The final kinetic energy of the system is expressed as;
KE = 1/2(m1+m2)v²
m1 and m2 are the masses of the two bodies
v is the final velocity of the bodies after collision
get the final velocity using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
0.12(0.45) + 0/12(0) = (0.12+0.12)v
0.054 = 0.24v
v = 0.054/0.24
v = 0.225m/s
Get the final kinetic energy;
KE = 1/2(m1+m2)v
KE = 1/2(0.12+0.12)(0.225)²
KE = 1/2(0.24)(0.050625)
KE = 0.12*0.050625
KE = 0.006075Joules
Hence the final kinetic energy of the system is 0.006075Joules
Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:
where
= Torque due to Force, F
= Torque due to Force, F'
Now,
The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque:
Answer:
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Explanation: