The acceleration is 3.3 m/s2
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that

For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration


T= I α
0.5= 0.25 α

For Wheel B
m= 1 kg
d= 2 m,r=1 m


Given that angular acceleration is same for both the wheel

T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N
The period of the pendulum is given by the following equation
T = 2π * sqrt (L/g)
Where g is the gravity (free fall acceleration)
L is the longitude of the pendulum
T is the period.
We find g.............> (T /2π)^2 = L/g
g = L/(T /2π)^2...........> g = 22.657 m/s^2
The answer is number 2 stomata.