Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Answer:
electricity because it has more percentage nd energy
Explanation:
mark me brainliest pl
Answer:(a) 50 N
(b)38.34 N
Explanation:
Given
Maximum tension(T) in line 50 N
(a)If line is moving up with constant velocity i.e. there is no acceleration
This will happen when Tension is equal to weight of Fish
T-mg=0
T=mg
Maximum weight in this case will be 50 N
(b)acceleration of magnitude 
T-mg=ma
m=3.91
Therefore weight is 
Answer:
46,800 seconds. I hope this helps
Answer:
Explanation:
velocity of first projectile after 3 s
v = u - gt
v = 49.4 - 9.8 x 3
= 20 m /s
Velocity of second projectile after 3 s after being dropped from rest
v = u + gt
= 0 + 9.8 x 3
= 29.4 m /s
They will be moving in opposite direction at the time of meeting , so their relative velocity
= 20 + 29.4 = 49.4 m /s
From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .
