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3 years ago

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A heat engine operates between 2 reservoirs at TH and 18oC. The heat engine receives 17,000 kJ/h from the high temperature reser

voir, and delivers half of its power output to drive a Carnot heat pump. The Carnot heat pump removes heat from the cold surroundings at 0oC and transfers it to a house which is maintained at 24oC. If the house is losing heat at a rate of 80,000 kJ/h, determine the temperature TH of the heat engine reservoir.

1 answer:

lisabon 2012 [21]3 years ago

4 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

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A 200 W vacuum cleaner is powered by an electric motor whose efficiency is 70%. (Note that the electric motor delivers 200 W of

Goshia [24]

Answer:

The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.

Explanation:

power efficiency of electric motor = 70% = 0.70

The power output of the vacuum cleaner = $P_o$ = 200 W

The power output of the vacuum cleaner = $P_i$

$Efficiency=\frac{P_o}{P_i}$

$0.70=\frac{200 W}{P_i}$

$P_i=\frac{200 W}{0.70}=285.71 W$

The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.

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Ben İngiliz oldum düzelte bilirmiyim

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Answer:

What laguange is that?

Explanation:

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3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

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What are the advantages and disadvantages of a mine heardgear

Irina18 [472]

Answer:

If there is a shaft with headgear, then mining can take place until that depth of the shaft. If it is accessed horizontal Adits, it can mine until the lowest Adit from upwards. If it is accessed decline, the development and mining can continue so long as economic exploitation is possible.

Explanation:

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Answer:

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Explanation:

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