Complete question:
A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.18 kg.m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?
Answer:
The angular velocity of the disk is 0.0464 rad/s
Explanation:
Given;
initial angular velocity of disk, ωi = 0.067 rad/s
initial moment of inertia of the disk, I₁ = 0.18 kg.m²
radius of sand on the disk, R = 0.40 m
mass of sand, m = 0.50 kg
Initial angular momentum = Final angular momentum

Moment of inertia of sand ring = MR²

Therefore, the angular velocity of the disk is 0.0464 rad/s
Answer:F=400.28 N
Explanation:
Given
mass of large cube 
mass of small cube 
coefficient of static friction 
acceleration a of the system is given by


Now Normal reaction between two blocks is given by

friction tries to balance weight
thus 
and 






Explanation:
here,load=100N
effort=200N
distance travelled by load=2cm
distance travelled by effort=10cm
Now, output work=L*Ld
=100*2
=200g
Again,
input work =E*Ed
200*10
2000j
again, efficiency = output work/input work*100%
200/2000*100%
=10%
please don't forget to write symbol
Answer:
159 N
Explanation:
The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then
Fr=0.25*636=159 N
Therefore, the force of friction is equivalent to 159 N