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Rom4ik [11]
3 years ago
9

(d) If η = 40% and TH = 427°C, what is TC, in °C?

Physics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

T_C=256.2^{\circ}C

Explanation:

Given that,

Efficiency of heat engine, \eta=40\%=0.4

Temperature of hot source, T_H=427^{\circ}C

We need to find the temperature of cold sink i.e. T_C. The efficiency of heat engine is given by :

\eta=1-\dfrac{T_C}{T_H}

T_C=(1-\eta)T_H

T_C=(1-0.4)\times 427

T_C=256.2^{\circ}C

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0
3 years ago
How many moles are in 73.4 grams of Phosphorus?
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20 ohms in parallel with 16 ohm= 8.89

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3 years ago
A wheel rotates clockwise 10 times per second. what is its angular speed answer
bogdanovich [222]
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