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Vadim26 [7]
2 years ago
10

Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here : 750 J electrical energy i

s transformed into 400 J kinectic energy . What happened to the remaining 350j of energy?
Physics
1 answer:
Charra [1.4K]2 years ago
3 0

Electrical energy is used to run the fan

Here as per given condition 750 J of electrical energy is used to run the fan which is converted into Kinetic energy as 400 J

So here we can see that 350 J of energy is lost against many other type of frictional and resistive loses.

So here we can say that out of 750 J of energy only 400 J is used to run the fan and rest amount of energy is lost against friction.

also we can say that efficiency of this fan will be

\eta = \frac{Useful}{Total}

\eta = \frac{400}{750}

\eta = \frac{8}{25}


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Answer:

D. 3 hours or more

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The average 8- to 18-year-old spends at least D. 3 hours every day in front of a screen, performing little to no physical activity. This is because, instead of exercising and socializing with their peers, children and teenagers frequently talk, watch a lot of movies/shows, or play video games on their computers. Unfortunately, this is typically considerably more than three hours every day. Although some children still prefer physical activities over this, the bulk of the population does not.

5 0
2 years ago
Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.
Gemiola [76]

Answer:

a)   i₈ = 0.5 i₄,  b)   i₁₀ = 0.3 i₃,    i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

       V = i R

        i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

         

R = 4 Ω

         i₄ = V / 4

R = 8 Ω

         i₈ = V / 8

we look for the relationship between these two currents

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R = 10 Ω

         

        i₁₀ = V / 10

   

we look for relationships

       i₁₀ / 1₃ = 3/10

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7 0
2 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

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∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

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Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

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