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gavmur [86]
3 years ago
11

Which of the following changes will always increase the efficiency of a thermodynamic engine? Choose all correct statements.

Physics
2 answers:
slava [35]3 years ago
6 0

Answer:

option (c)

Explanation:

The efficiency of the engine is given by

\eta =1-\frac{T_{c}}{T_{H}}

to increase the efficiency of the engine, the temperature of the cold reservoir should be decreased and thus the temperature of hot reservoir be increased.

So, decrease the ratio of Tc/TH.

option (C)

Ilya [14]3 years ago
5 0

Answer:B,C,D

Explanation:

Thermodynamic  efficiency is given by

\eta =1-\frac{T_C}{T-H}

\eta efficiency can be increased by Keeping _c constant and increasing T_H

Keeping T_H constant and decreasing T_c

by increasing \Delta T=T_H-T_c

by decreasing \frac{T_C}{T_H} ratio          

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A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter
tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

<em>F = Mg</em>

<em>⇒F = 84.8×9.81</em>

<em>⇒F = 831.888 N</em>

Area of the contact point

<em>A = πR²</em>

<em>⇒A = π0.006²</em>

<em>⇒A = π0.000036 m²</em>

Area of the four points is

<em>4A = 0.000144π m²</em>

Pressure

p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0
3 years ago
Convert -13°F into (a) °C (b) kelvin​
RideAnS [48]

Answer:

-25ºC

Explanation:

3 0
2 years ago
A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player
White raven [17]

Answer:

(a) -472.305  J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by0.5mv^{2}

Initial kinetic energy is 0.5\times 69\times 3.7^{2}=472.305 J

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305  J=-472.305  J

(b)

From fundamental kinematic equation

v^{2}=u^{2}+2as

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

s=\frac {v^{2}-u^{2}}{-2a} but a=\mu g hence

s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m

7 0
3 years ago
PLEASE HELP !!!
elixir [45]

Answer:

A

Explanation:

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