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Alex777 [14]
3 years ago
14

After running for a while, the fan becomes hot to the touch, and ever so often releases ___________________ energy into the air.

Physics
1 answer:
kykrilka [37]3 years ago
4 0

Answer:

thermal

Explanation:

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With all his gear, Neil Armstrong weighed 360 pounds on Earth. When he landed on the Moon, he weighed 60 pounds. Why?
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A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4
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Answer:

Explanation:

Let v is the launch speed of the plastic ball and the angle of projection is θ.

So, in horizontal direction

v Cosθ x t = 4.8 .... (1)

In th evertical direction

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t = 0.7 s

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v Cosθ x 0.7  = 4.8

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and v Sinθ x 0.7 = 3.8

v Sinθ = 5.43

Now

v^{2}\left ( Sin^{2}\theta +Cos^{2}\theta  \right )=5.43^{2}+6.86^{2}

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How’s does my friend cell phone communicate with them? How does it speak with my voice to them?
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1 year ago
An external resistor with resistance R is connected to a battery that has an emf E and an internal resistance r. Let P be the el
satela [25.4K]

Answer:

a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes

Explanation:

<u>Solution  :</u>

(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation  in the next form  

P=∈*I-I^2*r                (1)

Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by  

I=∈/R+r                     (2)

When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

I= ∈/r

Now let us plug this expression of I into equation (1) to get the consumed power  

P=∈*I-I^2*r

 =I(∈-I*r)

 =0

The consumed power when R is very small is zero  

(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes  

I=∈/R

The dissipated power due toll could be calculated by using equation.

P=I^2*r                (3)

Now let us plug the expression of I into equation (3) to get P  

P=I^2*R=(∈/R)^2*R

 =∈^2/R

4 0
3 years ago
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