A heat pump operates on a Carnot heat pump cycle with a COP of 8.7. It keeps a space at 248C by consum-ing 2.15 kW of power. Det
1 answer:
Answer:
<em>Heat of the reservoir is 461.38 K or 188.1 °C</em>
<em>The heating load is 18.705 kW</em>
Explanation:
COP = 8.7
working temperature = 248 °C = 248 + 273.3 = 521.3 K
work power W = 2.15 kW
reservoir temperature = ?
heating load Q = ?
We know that
COP = Q/W
Q = COP x W = 8.7 x 2.15 = <em>18.705 kW</em>
Also,
COP = =
8.7 =
4535.31 - 8.7 = 521.3
4535.31 - 521.3 = 8.7
4014.01 = 8.7
= 4014.01/8.7 = <em>461.38 K</em>
or 461.38 -273.3 = <em>188.1 °C</em>
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Answer:
Tension in right wire = 25.9N
Explanation:
I have attached a free body diagram to depict this question.
From the diagram, i have labelled the tensions in the strings T1 and T2.
While i labelled the weight of the bar as Wb and weight of sausage as Ws.
Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)
Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.
Thus;
T1 + T2 = Wb + Ws
We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N
Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N
Thus,
T1 + T2 = 48.46N + 24.43
T1 + T2 = 72.89N - - - - - (eq 1)
Now, let's take moments about the left end of the bar.
The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m
So, moments about left end;
T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)
1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)
1.46T2 = 35.373 + 2.443
1.46T2 = 37.816
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Answer:
a) &
b)
c)
Explanation:
Given:
mass of the book,
combined mass of the student and the skateboard,
initial velocity of the book,
angle of projection of the book from the horizontal,
a)
velocity of the student before throwing the book:
Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.
where:
initial velocity of the student
velocity of the student after throwing the book:
Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.
where:
final velcotiy of the student after throwing the book
b)
c)
Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.