Answer:

Explanation:
The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.
The above formula can be rewritten as follows

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.
If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.
Therefore the electric flux can be written as a function of time

Answer:
Impedance increases for frequencies below resonance and decreases for the frequencies above resonance
Explanation:
See attached file
Explanation:
Answer:

Explanation:
The attached figure shows the whole description. Considering the applied force is 100 N.
The acceleration of both blocks A and B, 
Firstly calculating the mass m using the second law of motion as :
F = ma
m is the mass


m = 125 kg
It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N



So, the new acceleration of the block is
. Hence, this is the required solution.
240 meters
Explanation: 16 x 15 because it’s 16 meters every second
Answer:
Explanation:
In a frictionless system with no acceleration, the tension in the rope must be F along its entire length
FBD analysis of the lower pulley has two upward acting tension vectors F and one downward acting weight vector W
2F = W
F = W/2
FBD analysis of the upper pulley has one upward acting support vector T and three downward acting tension vectors F
T = 3F
T = 3(W/2)
T = 1.5W