Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
Answer:
ω = 4.07 rad/s
Explanation:
By conservation of the energy:
W = ΔK

where 
Solving for ω:

Answer:
99.63 kg
Explanation:
From the force diagram
N = normal force on the worker from the surface of the roof
f = static frictional force = 560 N
θ = angle of the slope = 35
m = mass of the worker
W = weight of the worker = mg
W Cosθ = Component of the weight of worker perpendicular to the surface of roof
W Sinθ = Component of the weight of worker parallel to the surface of roof
From the force diagram, for the worker not to slip, force equation must be
W Sinθ = f
mg Sinθ = f
m (9.8) Sin35 = 560
m = 99.63 kg
Answer:
The angular speed of the wheel is 0.452 rad/s
Explanation:
The angle through which the car wheel turns, Δθ = 277° = 277/360 × 2·π radian
The time it takes for the car wheel to turn, Δt = 10.7 s
The angular speed, ω is given by the following equation;

Substituting the known values for Δθ and Δt gives;

The angular speed of the wheel = 0.452 rad/s
Answer:
B. i=2.79A
C. F=0.066N
Explanation:
A) By the right hand rule we have that
F=iL x B
F=iLBsin(α)
If the wire jump toward the observer the top pole face is the magnetic southpole.
B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

C) Now the length of the wire that feels B is

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

I hope this is useful for you
regards