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2 years ago

A perpetual-motion machine can never be built because it is not possible to eliminate...

Physics

2 answers:

bazaltina [42]2 years ago

4 0

The answer is: [C]: "elasticity" .
________________________________________

kykrilka [37]2 years ago

4 0

Answer:

D) Friction

Explanation:

Perpetual Motion machine is that machine which will convert whole input energy into useful work.

So we will say that perpetual motion machine is an hypothetical machine which will be 100% efficient and it will convert whole input energy into useful work without any loss of energy.

we know that efficiency is given as

$efficiency = \frac{output}{Input}$

So in real conditions it is not possible to eliminate frictional force due to which we will not able to get 100% useful work so here in real conditions the efficiency is always less than 100%

so correct answer will be

D) Friction

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f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an

vazorg [7]

Answer:

The magnitude of the angular acceleration is $a = 20.14 rad/s^2$

Explanation:

From the question we are told that

The angular speed of CD is $w_{CD} = 500 rpm = \frac{500 rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s$

time taken to decelerate is $t_{CD} = 2.60\ s$

The final angular speed is $w_f= 0 \ rad/s$

The angular acceleration is mathematically represented as

$a = \frac{w_f - w_{CD}}{t}$

substituting values

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3 0

3 years ago

Need help please!!!!

Levart [38]

I think it would be d because development in nations needs more population

5 0

3 years ago

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Evgesh-ka [11]

Answer:

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2 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

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3 years ago

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mart [117]

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3 years ago