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gayaneshka [121]
3 years ago
10

An engine has a hot-reservoir temperature of 980 K and a cold-reservoir temperature of 570 K. The engine operates at three-fifth

s maximum efficiency. What is the efficiency of the engine?
Physics
2 answers:
hram777 [196]3 years ago
7 0

Answer:

efficiency of engine is 0.21

Explanation:

given:

Th=980 K

Tc=570 K

since efficiency of engine is given as:

n=3/5*n_m

 =3/5(1-Tc/Th)

 =0.21

efficiency of engine is 0.21

pshichka [43]3 years ago
4 0

Answer:

E = 0.251 or 25.1%

The efficiency of the engine is 0.251 or 25.1%

Explanation:

Maximum Efficiency of an engine can be expressed as;

Em = workdone/heat energy supplied

Em = 1 - Qc/Qh = 1 - Tc/Th

Given;

Tc = cold temperature = 570K

Th = hot temperature = 980K

Substituting the values;

Em = 1 - 570/980

Em = 0.4184

But the engine operates at three-fifths the maximum efficiency

E = 3/5 × Em

Substituting Em

E = 3/5 × 0.4184

E = 0.251 or 25.1%

The efficiency of the engine is 0.251 or 25.1%

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Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

\frac{\Delta l}{l_{o}}=\frac{F}{AY} (1)

T=2 \pi \sqrt{\frac{l_{o}}{g}} (2)

Where:

\Delta l=0.05 mm=5(10)^{-5} m is the length the steel wire streches (taking into account 1mm=0.001 m)

l_{o} is the length of the steel wire before being streched

F=mg=(2 kg)(9.8 m/s^{2})=19.6 N is the force due gravity (the weight) acting on the pendulum with mass m=2 kg

A is the transversal area of the wire

Y=2(10)^{11} Pa is the Young modulus for steel

T is the period of the pendulum

g=9.8 m/s^{2} is the acceleration due gravity

Knowing this, let's begin by finding A:

A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4} (3)

Where d=1.1 mm=0.0011 m is the diameter of the wire

A=\pi \frac{(0.0011 m)^{2}}{4} (4)

A=9.5(10)^{-7}m^{2} (5)

Knowing this area we can isolate l_{o} from (1):

l_{o}=\frac{\Delta l AY}{F} (6)

And substitute l_{o} in (2):

T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}} (7)

T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}} (8)

Finally:

T=1.39 s

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Option D.

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