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natka813 [3]
3 years ago
14

An internal combustion engine does 356 kJ of useful work using 946 kJ of thermal energy from the gasoline consumed in the engine

. What is the efficiency of this engine?
Physics
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

The efficiency of this engine is 37.63 %.

Explanation:

Given;

useful output work done by the combustion engine, = 356 kJ

input work, = 946 kJ

The efficiency of the combustion engine is calculated as;

Efficiency = \frac{0utput \ work}{1nput \ work} \times 100\%\\\\Efficiency = \frac{356 \ \times \ 10^3}{946 \ \times \ 10^3} \times 100\%\\\\Efficiency = 37.63 \ \%

Therefore, the efficiency of this engine is 37.63 %.

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A manometer using oil (density 0.900 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases
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Answer:

Rise in level of fluid is 0.11 m

Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm

Solution:

As per the question:

Density of oil, \rho_{o} = 0.900\ g/cm^{3} = 900\ kg/m^{3}

Change in Pressure in the tank, \Delta P = 7.28\ mmHg

Density of the mercury, \rho_{m} = 13.6\ g/cm^{3} = 13600\ kg/m^{3}

Now,

To calculate the rise in the level of fluid inside the manometer:

We know that:

1 mmHg = 133.332 Pa

Thus

\Delta P = 7.28\ times 133.332 = 970.656\ Pa

Also,

\Delta P = \rho_{o} gh

where

g = acceleration due to gravity

h = height of the fluid level

970.656 = 900\times 9.8\times h

h = 0.11 m

Now, if mercury is used:

\Delta P = \rho_{m} gh

970.656 = 13600\times 9.8\times h

h = 0.00728 m = 7.28 mm

3 0
4 years ago
Two vectors are illustrated in the coordinate plane.
oksano4ka [1.4K]

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We can write this in terms of its components:

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After completing the group activity, do you believe that your instructions were written enough for the other team to follow?
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Answer:

yes I do

Explanation:

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