Answer:
False
Explanation:
Faraday's law gives the relationship between the induced emf and the rate of change of magnetic flux i.e.

The given statement "A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change" is false. The reason is that if the rate of change of magnetic flux is greater, then its will induce more emf. It would mean it does not say about emf.
Hence, it is false.
Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:

where
= Torque due to Force, F
= Torque due to Force, F'


Now,


The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque:

Answer:
The magnitude of the electric flux is 
Explanation:
Given that,
Electric field = 2.35 V/m
Angle = 25.0°
Area 
We need to calculate the flux
Using formula of the magnetic flux


Where,
A = area
E = electric field
Put the value into the formula



Hence, The magnitude of the electric flux is 
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
<em>Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.</em>
<em>You are correct! **</em>