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2 years ago

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A Carnot engine receives 250 kJ·s−1 of heat from a heat-source reservoir at 525°C and rejects heat to a heat-sink reservoir at 5

0°C. What are the power developed and the heat rejected?

1 answer:

earnstyle [38]2 years ago

5 0

Answer:

W = -148.8 kJ/s

Qc= -101.2 kJ/s

Explanation:

<u>note: </u>

<u> Solution is attached in word form due to error in mathematical equation. please find the attachment </u>

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Ideally, rewards should be given immediately and frequently but

snow_lady [41]

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When are you most aware of your motion in a moving vehicle: when it is moving steadily in a straight line or when it is accelera

Ghella [55]

Answer:

Acceleration is percieved, not constant velocity.

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It's because of this that:

(A) False. The acceleration can be detected from the inside of a closed car.

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2 years ago

While i was jogging, i wad able to run 5 miles in thirty minutes. from this information, we can determine my?

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2 years ago

Read 2 more answers

If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m

algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

$H=ut+\dfrac{1}{2}at^2$

Here, a = g

$H=\dfrac{1}{2}gt^2$

$H=\dfrac{1}{2}\times 9.8\times (2.261)^2$

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

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2 years ago

A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw

insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

$[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a$ which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

$[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a$

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

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3 years ago