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2 years ago

A computer that is 87% efficient consumes 375 kWh of energy. How much useful energy does it provide?

Physics

1 answer:

tekilochka [14]2 years ago

3 0

Answer:

326.25 kWh

Explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}$

Plug in the values of $\eta=0.87$ and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}$

Therefore, the useful energy provided by the computer is 326.25 kWh.

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

2 years ago

A man stands on his balcony, 130 feet above the ground. He looks at the ground, with his sight line forming an angle of 70° with

jenyasd209 [6]

Answer:

d = 380 feet

Explanation:

Height of man = perpendicular= 130 feet

Angle of depression = ∅ = 70 °

distance to bus stop from man = hypotenuse = d = 130 sec∅

As sec ∅ = 1 / cos∅

so d = 130 sec∅ or d = 130 / cos∅

d = 130 / cos(70°)

d = 380 feet

8 0

2 years ago

Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air

vova2212 [387]

Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

The mass is equal to:
Mass = (0.00118g/cm³)(0.0593 cm³)
Mass = 7 grams of HCN

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2 years ago

Which of the following exists around every object that has mass?

ICE Princess25 [194]

Answer:

A is the correct answer.

Explanation:

6 0

2 years ago

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The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c

puteri [66]