Answer:
The average emf induced in the coil is 175 mV
Explanation:
Given;
number of turns of the coil, N = 1060 turns
diameter of the coil, d = 20.0 cm = 0.2 m
magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T
duration of change in field, t = 10 ms = 10 x 10⁻³ s
The average emf induced in the coil is given by;
where;
A is the area of the coil
A = πr²
r is the radius of the coil = 0.2 /2 = 0.1 m
A = π(0.1)² = 0.03142 m²
Therefore, the average emf induced in the coil is 175 mV
Answer:
W = 1.432 KJ
Explanation:
given,
mass = 22.2 Kg
angle of the rope = 27.5°
distance on the ground = 24 m
kinetic friction= μ = 0.32
acceleration due to gravity, g = 9.8 m/s²
Work done = ?
W = F d cosθ
a = 0 because it is moving with constant speed
equating all the forces acting in x direction
F cosθ = F friction = μN
equating all the forces acting in y direction
F sinθ + N -mg =0
now,
N = mg - F sinθ
putting value of N
F cosθ = μ mg -μ F sinθ
F (cosθ + μsinθ ) = μ mg
F =67.28 N
now,
W=F d cosθ
W =67.28 x 24 x cos(27.5)
W =1432.27 J
W = 1.432 KJ