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1 year ago

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

Physics

1 answer:

tekilochka [14]1 year 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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1 year ago

A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of th

jekas [21]

Answer:

clockwise

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Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.

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1 year ago

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

andrey2020 [161]

Answer:

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1 year ago

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

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- We use the relation between the intensity I and distance of light:

I_1 / I_2 = ( r_2 / r_1 ) ^2

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1 year ago

Find the kinetic of a 0.1 kilogram toy truck moving at a speed of 1.1 meters per second

omeli [17]

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KE = (0.5) times the mass times the velocity^2
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square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
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remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47

GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height

height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
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