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:
the exposed core of a dead star, supported by electron degeneracy pressure.
Explanation:
A white dwarf is a low luminosity exposed core of a dead star having mass comparable to the sun but volume comparable to the earth . So its density is very high . These stars have lost the capacity to generate energy through the process of fusion . Due to high gravitational energy , it goes on shrinking but ultimately balanced by electron degeneracy pressure. It is not a main sequence star as it has lost the power of fusion .
Answer:
0.001 s
Explanation:
The force applied on an object is equal to the rate of change of momentum of the object:

where
F is the force applied
is the change in momentum
is the time interval
The change in momentum can be written as

where
m is the mass
v is the final velocity
u is the initial velocity
So the original equation can be written as

In this problem:
m = 5 kg is the mass of the fist
u = 9 m/s is the initial velocity
v = 0 is the final velocity
F = -45,000 N is the force applied (negative because its direction is opposite to the motion)
Therefore, we can re-arrange the equation to solve for the time:

Our data are,
State 1:

State 2:

We know as well that 
To find the mass we apply the ideal gas formula, which is given by

Re-arrange for m,

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

Replacing,

For conservative energy we have, (Cv = 0.718)

Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec