Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = 
- d fi / dt
= B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A 
Dt = - A 
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
Answer:
The frictional force acting on the block is 14.8 N.
Explanation:
Given that,
Weight of block = 37 N
Coefficients of static = 0.8
Kinetic friction = 0.4
Tension = 24 N
We need to calculate the maximum friction force
Using formula of friction force
Put the value into the formula
So, the tension must exceeds 29.6 N for the block to move
We need to calculate the frictional force acting on the block
Using formula of frictional force
Put the value in to the formula
Hence, The frictional force acting on the block is 14.8 N.
Answer:
Explanation:
You can calculate the total electric charge that passes through the conductor as 
. It means that the number of electron that passes through the conductor is:
Um student a because they were there a few seconds ahead
In a constant acceleration of 3m per second, after 10 seconds,
3 x 10 = 30
B. 30m/s is your answer
hope this helps :D
