Two solid disks, which are free to rotate independently about the same axis that passes through their centers and perpendicular to their faces, are initially at rest. The two disks have the same mass, but one of has a radius R and the other has a radius 3R. A force of magnitude F is applied to the edge of the smaller radius disk and it begins rotating. What force must be applied to the edge of the larger disk so that the angular acceleration is the same as that for the larger disk

3F

Explanation:

given

radius of disk 1 r_1 = R

radius of disk 2 r_2 = 3 R

Also, force applied to smaller radius disk is f1 = F

moment of inertia of small disk

I = 1/2MR^2

torque produce

T = R×F

Now, angular acceleration α= T÷I = R×F/0.5MR^2

given that angular acceleration for both disks is same .

Therefore,

α_1 = α_2

Thus,

R×F/0.5×M×R2 = 3R×F'/0.5×M×(3R)2

Solving we get

F' = 3×F

b.   $146

Step-by-step explanation:

Given: The fixed parking charge in the lot = $2

The number of hours of parking =72 hours

The number of half hours in 1 hour =2

The number of half hours in 72 =

Since, they charge $1 per half-hour parked, the additional charge will be given by :-

Hence, the  total charge from parking in the lot for 72 hours=$2+$144=$146.