Answer:
2.5 m
Explanation:
Weight of billboard worker = 800 N
Number of ropes = 2
Length of scaffold = 4 m
Weight of scaffold = 500 N
Tension in rope = 550 N
The sum of the torques will be

The position of the person will be 2.5 m
Answer:
Explanation:
Given
Diameter of Pulley=10.4 cm
mass of Pulley(m)=2.3 kg
mass of book
height(h)=1 m
time taken=0.64 s


![a=4.88 m/s^2and [tex]a=\alpha r](https://tex.z-dn.net/?f=a%3D4.88%20m%2Fs%5E2%3C%2Fp%3E%3Cp%3Eand%20%5Btex%5Da%3D%5Calpha%20r)
where
is angular acceleration of pulley


And Tension in Rope


T=8.364 N
and Tension will provide Torque




Thus mass is uniformly distributed or some more towards periphery of Pulley
A-200 kg
I hope this helped xx
Answer:
5.38 m/s^2
Explanation:
NET force causing the object to accelerate = 50 -10 = 40 N
Mass of the object = 73 N / 9.81 m/s^2 = 7.44 kg
F = ma
40 = 7.44 * a a = 5.38 m/s^2