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djyliett [7]
3 years ago
11

An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?

Physics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%

so, the efficiency of heat engine is 13.7%.

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Sever21 [200]

Answer:

minimum mass of the neutron star = 1.624 × 10^30 kg

Explanation:

For  a material to remain on the surface of a rapidly rotating neuron star, the magnitude oĺf the gravitational acceleration on the material must be equal to the magnitude of the centripetal acceleration of the rotating neuron star.

This can be represented by the explanations in the attached document.

minimum mass of the neutron star = 1.624 × 10^30 kg

8 0
3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

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We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

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3 years ago
What is number 14 please tell me
UNO [17]
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Which statement best describes how the temperature of the oceans surface water varies
lorasvet [3.4K]
If the temperature is high there is less water because it evaporates if it is cloudy it is more because it doesn't evaporate
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A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the
valina [46]
The speed of water can be split into vertical and horizontal speed components:
v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta

Due to the force of gravity, the y component will be parabolic. The x component will be linear:
y(t) = -4.9t^2 + (6.5sin \theta) t \\  \\ x(t) = (6.5 cos \theta) t
To find when the water hits the ground 2.5m away, set y= 0 and x = 2.5
-4.9t^2 + (6.5sin \theta) t=0 \\  \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\  \\ sin \theta cos \theta = 0.29 \\  \\ sin 2\theta = 0.58 \\  \\ 2\theta = 35.4, 144.6 \\  \\ \theta = 17.7,72.3
8 0
3 years ago
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