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;
Answer:
6 km is the right answer
hope it helps you
Explanation:
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Where's the person that the question is talking about?
Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
