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amm1812
4 years ago
15

A system undergoing a power cycle develops a steady power output of 0.3 kW while receiving energy input by heat transfer at a ra

te of 2400 Btu/h. Determine the thermal efficiency and the total amount of energy developed by work, kW · h, for 1 full year of operation.
Physics
1 answer:
uranmaximum [27]4 years ago
4 0

Answer:

The thermal efficiency is 43%.

The total amount of energy developed by work is 397 kW in one year.

Explanation:

Given that,

Power output = 0.3 kW

Heat energy = 2400 Btu/h = 0.703 kW

We need to calculate the thermal efficiency

Using formula of efficiency

\eta=\dfrac{out-put}{in-put}

Put the value into the formula

\eta=\dfrac{0.3}{0.703}

\eta=0.43\times100

\eta=43\%

The thermal efficiency is 43%.

(b). We need to calculate the total amount of energy

Using formula of energy

E=0.43\times0.703\times60\times60\times365

E=397209.06\ J

E=397\ kW

Hence, The thermal efficiency is 43%.

The total amount of energy developed by work is 397 kW in one year.

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Which of these biomes receives the least amount of rainfall every year?
ASHA 777 [7]

Answer:

Desert

Explanation:

The desert receives the least amount of rainfall yearly.

A desert can be described a barren area of land where little amount of rainfall occurs and, because of this, living conditions are not favourable for plants and animal life. Because there is no vegetation in a desert, the bare surface of the ground are subjected to denudation.

5 0
3 years ago
You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
wlad13 [49]

Answer:

  • between locations that are 14 cm outboard of the chair edges
  • the weightless board is centered and end sitters are 25 cm from the ends

Explanation:

We can assume the .5 m-wide chair means that it is comfortable for each student to sit 0.25 m from the end of the board. If the board is centered on the chair, then each student is 1 m from the edge of the chair.

When Dan and Tahreen are seated on the board, their center of mass is ...

  (50 kg×2.5 m)/(50 kg +67 kt) = 1.068 m

to the right of the position where Dan is seated. Since this location is over the chair, the board is stable.

Komila can sit as much as x distance from the chair toward Dan, where ...

  67(1) +54(x) = 50(1.5)

  x = 8/54 ≈ 0.148 . . . . meters

Or, Komila can sit as much as x distance from the chair toward Tahreen, where ...

  67(1.5) = 54(x) +50(1)

  x = 50.5/54 ≈ 0.935 . . . . meters

<u>Scenario 1</u>

Assuming the (weightless) board is centered on the chair, Komila can sit anywhere between 14.8 cm left of the chair and 93.5 cm right of the chair and the board will remain stable. Sitting on the board centered on the chair is a suitable location. The two students sitting on the ends must become (and stay) seated at the same time. They both must be seated 0.25 m from the end of the board for the other dimensions to remain valid.

<u>Scenario 2</u>

Assuming the (weightless) board is located so its left end is 1.068 m from the chair, and Dan and Tahreen are seated 0.25 m from the ends of the board, Komila can sit anywhere within (117/54×.25 m) = 0.54 m of the chair and the board will remain stable. Again, sitting centered on the chair is a suitable location.

__

There does not appear to be any location where Komila can sit and have the board remain stable with only Dan or Tahreen seated on one end (assuming a width of 0.5 m for each sitter).

_____

<em>Comment on the question</em>

For the board to remain stable, the sum of moments about either edge of the chair must tend to rotate the board toward the chair. This sum will depend on the locations of the sitters relative to each edge of the chair, so there is significant freedom in choosing locations. To make the problem tractable, we have made some specific assumptions about where the board is and what the locations of the sitters might be. YMMV

3 0
4 years ago
PLEASSEE HELP MEEE WITHH THESEEE
DanielleElmas [232]

Answer:

Q1: (a)

1cm \:  = 10 {}^{ - 2}m

so

45 \times  {10}^{ - 2} m

Q2:

1km = 10 {}^{3} m \\ 6378km \times  {10}^{ 3}

this is the radius in meter

diameter = 2r

in this i don't know which choice is the right answer

Q3:

1gram = \:  {10}^{ - 3} kg

so

250 \times  {10}^{ - 3}  = 2.5 \times  {10}^{ - 1} kg

again there is no correct answer

7 0
3 years ago
How is the mass number calculated for an element?
natali 33 [55]
<span># of protons + # of neutrons = atomic mass</span>
8 0
4 years ago
The tortoise and the hare are having a race over a course of length 0.52 km. The tortoise crawls along at a constant 0.075 m/s w
EastWind [94]

Answer:the hare slept for 1 hr 55minutes.

Explanation: distance is changed to meters = 520m

Speed=distance/time

Speed of tortoise is 0.075m/s=520m/t

Crops multiplying t= 520/0.075 = 6933.3 seconds.

Speed for hare 16.7m/s =260m/t

ie 260m is half of the distance.

t =15.56seconds

Therefore the time slept by the hare = 6933.3- 15.56= 6917.74 seconds.

Converting to hours = 1hr. 55minutes is the time the hare slept.

3 0
3 years ago
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