Answer:
(a) The self inductance, L = 21.95 mH
(b) The energy stored, E = 4.84 J
(c) the time, t = 0.154 s
Explanation:
(a) Self inductance is calculated as;
where;
N is the number of turns = 1000 loops
μ is the permeability of free space = 4π x 10⁻⁷ H/m
l is the length of the inductor, = 45 cm = 0.45 m
A is the area of the inductor (given diameter = 10 cm = 0.1 m)
(b) The energy stored in the inductor when 21 A current ;
(c) time it can be turned off if the induced emf cannot exceed 3.0 V;
The answer is C in this question.
Answer:
250N
Explanation:
Given parameters:
Time = 4s
Momentum = 1000kgm/s
Unknown:
Force = ?
Solution:
To solve this problem, we use Newton's second law of motion;
Ft = Momentum
F is the force
t is the time
So;
F x 4 = 1000kgm/s
F = 250N
Answer:
- 0.6
Explanation:
Given that angle between normal y axis is 62° so angle between normal
and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence
Angle of incidence = 28 °
angle of reflection = 28°
Angle between incident ray and reflected ray = 28 + 28 = 56 °
Angle between x axis and reflected ray = 56 °
x component of reflected ray
= - cos 56 ( it will be towards - ve x axis. )
- 0.6
