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2 years ago

If something is 50% efficient, how many joules of wasted energy will there be if 750J of energy is put in?

Physics

2 answers:

Sidana [21]2 years ago

8 0

$\large{{\bold{ \underline{ \underline{ \overline{ \overline{ \red{☆Solution :—}}}}}}}}$

↪Total energy = 750 J

↪Efficiency = 50%

↪Wasted energy = 50% [ 100% - 50%]

Amount of wasted energy

↦ 750 x 50%

↦ 750 x 0.5

↦ 375 J

Andreas93 [3]2 years ago

7 0

Answer:

375 J

Explanation:

Total energy = 750 J

Efficiency = 50%

Wasted energy = 50% [100% - 50%]

Amount of wasted energy

= 750 x 50%

= 750 x 0.5

= 375 J

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Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

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Given:

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mass of fuel, $m_f=100\ kg$

acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

time after which the fuel exhaust, $t_f=35\ s$

During the phase of fuel exhaustion:

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$u=$ initial velocity of the rocket = 0

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$v_f^2=u^2+2a.h_f$

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$h_f=18375\ m$

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Time taken to reach the top height after the fuel is over:

$v=v_f+g.t'$

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Height ascended by the rocket after the fuel is over:

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

Therefore the maximum altitude attained by the rocket:

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

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$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$

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