Answer:
Efficiency = 52%
Explanation:
Given:
First stage
heat absorbed, Q₁ at temperature T₁ = 500 K
Heat released, Q₂ at temperature T₂ = 430 K
and the work done is W₁
Second stage
Heat released, Q₂ at temperature T₂ = 430 K
Heat released, Q₃ at temperature T₃ = 240 K
and the work done is W₂
Total work done, W = W₁ + W₂
Now,
The efficiency is given as:

or
Work done = change in heat
thus,
W₁ = Q₁ - Q₂
W₂ = Q₂ - Q₃
Thus,

or

or

also,

or

thus,

thus,

or

or
Efficiency = 52%
Answer:
P(bat) = V²r/(R+r)²
Explanation:
Let the resistance of the coil be R
Internal resistance of the battery be r
Emf of the battery = V
Power dissipated in the internal resistance of the battery is normally given as P = I²r
where I is the current flowing in the circuit.
From Ohm's law,
V = I R(eq)
R(eq) = (R + r)
I = V/(R+r)
P = I²r
P = [V/(R+r)]²r
P = V²r/(R+r)²
Hope this Helps!!!
If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.
The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.
- A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m
For more information on work done, visit
brainly.com/subject/physics