Answer:
Explanation:
Lets take the numerator of the fraction to be = x
So the denominator of the fraction is 4 more than the numerator = x+4
The fraction is ;
Now add 4 to the numerator and add 7 to the denominator as;
This new fraction is equal to 1 half =1/2
write the equation as;
perform cross-product
2(x+4 )=1( x+11 )
2x+8 = x + 11
2x-x = 11-8
x=3
The original fraction is;
Answer:
I=9.6×e^{-8} A
Explanation:
The magnetic field inside the solenoid.
B=I*500*muy0/0.3=2.1×e ^-3×I.
so the total flux go through the square loop.
B×π×r^2=I×2.1×e^-3π×0.025^2
=4.11×e^-6×I
we have that
(flux)'= -U
so differentiating flux we get
so the inducted emf in the loop.
U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)
so, I=2.9×e^{-6}÷30
I=9.6×e^{-8} A
The equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.
<h3>
What is resistance?</h3>
Resistance is the obstruction offered whenever the current is flowing through the circuit.
So the equivalent resistance is when three resistances are connected in series. When the resistances are connected in series then the voltage is different and the current remain same in each resistance.
V eq = V₁ + V₂ + V₃
IReq = IR₁ + IR₂ + IR₃
Req = R₁ + R₂ + R₃
Therefore the equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.
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Answer:
YES
Explanation:
Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit