Answer:
a) The schematic illustrating is attached
b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ
c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ
Explanation:
b) The heat transfer to the engine and the heat transfer from the engine to the air is:
Where
W = 750 kJ
n = 35% = 0.25
Replacing:
c) The efficiency of Carnot engine is:
The heat transfer to the heat engine is:
The heat transfer from the heat engine is:
The value of the peak power will be 50 W. The peak power is found as two times of the average power. Option C is correct.
<h3 /><h3>What is average power?</h3>
Average power is a ratio of total work done by the body to the total time. Its unit is the watt. It is also defined as the rate of total work done by the body.
The given data in the problem is;
Average power = 25 W
Peak power = ?
The peak power is found as;
Peak power = 2 × average power
Peak power=2×25
Peak power = 50 W
Hence, the value of the peak power will be 50 W.
To learn more about the average power, refer to the link;
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Answer:
a. El peso = 686 Newton
b. El peso = 113.4 Newton
Explanation:
Dados los siguientes datos;
Masa = 70 kg
Aceleración debida a la gravedad en la luna = 1,62 m/s² una.
a. Para encontrar la fuerza-peso en la Tierra;
Sabemos que la aceleración debida a la gravedad es igual a 9,8 m/s² en el planeta Tierra.
El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.
Matemáticamente, el peso de un objeto viene dado por la fórmula;
Dónde;
m es la masa del objeto.
g es la aceleración debida a la gravedad.
Substituting into the formula, we have;
El peso = 70 * 9.8
El peso = 686 Newton
b. To find weight on moon;
Weight = mass * acceleration due to gravity on moon
Weight = 70 * 1.62
Weight = 113.4 Newton
Answer:
a) 0.09N b) positive x direction
Explanation:
Force on a conductor carrying current in a magnetic field can be expressed as;
F = BILsin(theta) where
F is the force on the conductor (wire)
B is the uniform magnetic field I'm Tesla = 1.8Tesla
I is the current in the wire = 5×10^-2A
L is the length of the wire = 1m
theta is the angle that the conductor make with the magnetic field = 90° (since the wire in the horizontal direction is perpendicular to the field acting upwards)
Substituting this value in the formula to get F we have;
F = 1.8×5×10^-2×1 × sin90°
F = 0.09N
The force on the wire is 0.09N
b) The direction of the force is in the positive x direction since the wire acts horizontally to the magnetic field.