Answer:
1121.7 × 10³⁰ photons per second
Explanation:
Data provided in the question:
Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W
Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz
Now,
P = 
here,
N is the number of photons
t is the time
E = energy = hf
h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s
Thus,
P =
=
[t = 1 s for per second]
or
550 × 10³ = 
or
550 = N × 4903.24 × 10⁻³⁴
or
N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%

W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000

<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer: the answer for the question is a
Explanation:
Answer:
25 V
Explanation:
It is convenient to use Kirchoff's current law (KCL), which tells you the sum of currents into a node is zero. The node of interest is the top left node.
The currents into it are ...
20 mA + (-5 -Vo)/(2kΩ) -(Vo/(5kΩ)) = 0
20 mA -2.5 mA = Vo(1/(2kΩ) +1/(5kΩ)) . . . . add the opposite of Vo terms
(17.5 mA)(10/7 kΩ) = Vo = 25 . . . volts . . . . divide by the coefficient of Vo
_____
You will notice that the equation resolves to what you would get if you drew the Norton equivalent of the voltage source with its 2k impedance. You have two current sources, one of +20 mA, and one of -2.5 mA supplying current to a load of 2k║5k = (10/7)kΩ. KCL tells you the total current into the node is equal to the current through that load (out of the node).