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ICE Princess25 [194]

1 year ago

9

The ______ number of a flow is defined as the ratio of the speed of flow to the speed of sound in the flowing fluid.

1 answer:

allsm [11]1 year ago

4 0

Answer:

The statement is true

Explanation:

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A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the a

prisoha [69]

Answer:

W = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)

we know that maximum possible efficiency is given as

$\eta = 1- \frac{T_L}{T_H}$

$\eta = 1 - \frac{ 293}{368}$

$\eta = 0.2038$

rate of heat transfer is given as

$Q_H = \dot m C_p \Delta T$

$Q_H = 0.2 * 4.18 8(95 - 20)$

$Q_H = 62.7 kW$

Maximuim power is given as

$W = \eta Q_H$

W = 0.2038 * 62.7

W = 12.8 KW

3 0

3 years ago

How much metal can be removed from a cracked drum to restore surface

Alisiya [41]

Answer:sc

Explanation:

sc

7 0

3 years ago

Read 2 more answers

Select the correct answer.

Ulleksa [173]

The answer is A. Immediately inform her colleague

4 0

3 years ago

Basic concepts surrounding electrical circuitry?

Elanso [62]

Hopefully that helps you out and is this for history or science?

3 0

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 $0.711v_0$ .

5 0

3 years ago