In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
1 answer:
Answer:
η = 40 %
Explanation:
Given that
Qa ,Heat addition= 1000 J
Qr,Heat rejection= 600 J
Work done ,W= 400 J
We know that ,efficiency of a engine given as
Now by putting the values in the above equation ,then we get
η = 0.4
The efficiency in percentage is given as
η = 0.4 x 100 %
η = 40 %
Therefore the answer will be 40%.
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The equilibrium condition allows finding the results for the forces of the system are
a) The free body diagram is in the attachment
b) The normal force is N = 737 N
c) The mass of the table is 10.2 kg
Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition
∑ F = 0
Where the bold letters indicate vectors, F is the external forces
a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.
b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force
N = N₁ + N₂ + N₃ + N₄₄
Let's apply the equilibrium condition
N - = 0
N =
N =
They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N
N = 65 9.8 + 100
N = 737 N
c) To calculate the mass of the table we use the relation
W = g
e = 10.2 kg
In conclusion using the equilibrium condition we can find the results for the forces are
a) The free body diagram is in the attachment
b) The normal force is N = 737 N
c) The mass of the table is 10.2 kg
Learn more here: brainly.com/question/19860811
Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
