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Fudgin [204]
4 years ago
6

How much power can a motor output if it does 25 J of work in 1 s? *

Physics
2 answers:
Klio2033 [76]4 years ago
5 0

Answer:

25 W

Explanation:

Power: This can be defined as the rate at which work is done. The S.I unit of power is Watt (W).

The expression of power is given as

Power = work output/time

P = W/t................ Equation 1

Given: W = 25 J. t = 1s

Substitute into equation 1

P = 25/1

P = 25 W

Hence the output power of the motor is 25 W

Sergio039 [100]4 years ago
3 0

Answer:

The power is 25 W

Explanation:

Given;

work done by motor output. E = 25 J

Time taken to complete this task, t = 1 s

Power is the amount of work done per unit time or energy per unit time.

The amount of power developed by the motor, can be calculated as follows using the formula below;

Power = \frac{Energy}{time} = \frac{25}{1} = 25 \ W

Therefore, the amount of power developed by the motor in the given time is 25 W.

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4 0
3 years ago
Can someone please help me:(
finlep [7]

Answer:

<em><u>Incomple</u></em><em><u>te</u></em><em><u> </u></em><em><u>ques</u></em><em><u>tion</u></em><em><u>!</u></em>

you have to provide mass to get acceleration

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8 0
3 years ago
Which refers to a diagram that shows thermal energy being released by objects?
liq [111]

Answer:

The correct option is thermogram

Explanation:

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8 0
3 years ago
Read 2 more answers
Assume you live in a Einsteinian universe with the speed of light being only 60 k /hr. First you see a bicyclist at rest and lat
Free_Kalibri [48]
<span>The answer is that, Mass, time, and length are the quantities that would change considering the speed of light as the limit. In Einstein's equations it depends on velocity as a fraction of the speed of light.

we can take example as,
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4 0
3 years ago
The work function for cesium is 1.96 eV. (a) Find the cutoff wavelength for the metal, (b) what is the maximum kinetic energy fo
romanna [79]

Answer:

Explanation:

λ = hc/¢

Where

h = the Plank constant 6.63 x 10-34 Is

C = 3.0×10^8

¢= 1.96eV

= (6.63×10^-34Js)×(3×10^8)÷( 1.96eV) × 1eV/1.6×10^-19J

= (1.989×10^-25)÷( 1.96eV)×1eV/1.6×10^-19J

= 6.342×10^-7m

B) maximum kinetic energy

= K=hf−ϕ ........1

ϕ = hc

Where

​ h = constant 6.63 x 10^-34Js

ϕ= 1.96eV

Recall

λ =425×10^-9m

f = frequency in Hz

f = c / λ

C = 3.0×10^8

f = 3.0×10^8 / 425×10^-9m

f = 0.000705Hz

From equation 1

K = (6.63 x 10^-34Js×0.000705Hz )- 6.63 x 10^-34Js×3.0×10^8

= 4.68×10^-37 - 1.989×10^-25

= - 1.98×10^-25J

7 0
3 years ago
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