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Rom4ik [11]
4 years ago
9

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

Physics
1 answer:
Brrunno [24]4 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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x = 6.94 m

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For this exercise we can find the speed at the bottom of the ramp using energy conservation

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            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

            Em_{f} = K = ½ m v²

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Let's calculate

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In the horizontal part we can use the relationship between work and the variation of kinetic energy

            W = ΔK

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Newton's second law

              N- W = 0

     

The equation for the friction is

               fr = μ N

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We replace

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6 0
4 years ago
Read 2 more answers
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Answers:

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b) The direction of the balloon is towards the radar station

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f'=\frac{V+V_{o}}{V-V_{s}} f  (1)

Where:

f=250,000 Hz is the actual frequency of the sound wave

f'=240,000 Hz is the "observed" frequency

V=340 m/s is the velocity of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which is the balloon

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}  (2)

V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}  (3)

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Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (t=4.8 s) given as data, we can find the distance:

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3 years ago
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Therefore, the banking angle of the road is 24.2⁰

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