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3 years ago

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Water flows at low speed through a circular tube with inside diameter of 2 in. A smoothly contoured body of 1.5 in. diameter is

held in the end of the tube where the water discharges to atmosphere. Neglect frictional effects and assume uniform velocity profiles at each section. Determine (a) the pressure measured by the gage and (b) the force required to hold the body.

1 answer:

Art [367]3 years ago

5 0

Answer:

Pressure = 11.38 psi

Force = 13.981 Ibf

Explanation:

Step by step solution is in the attached document.

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Effects of adding more insulation to a cylinder increases heat transfer area. a)-True b)-False

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Answer:

a)-True

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Drivers killed in speed related accidents usually have a history of_______

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For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

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Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

$c=\sqrt(\gamma*R*T)$

Where R is the gas's particular constant defined in terms of Cp and Cv:

$R=Cp-Cv$

For particular values given:

$R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}$

$R=188.9 \frac{J}{Kg-K}$

The gamma undimensional constant is also expressed as a function of Cv and Cp:

$\gamma=Cp/Cv$

$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

$c=313 m/s$

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3 years ago

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Answer:

$\triangle V_0=0.08V$

Explanation:

From the question we are told that:

Incremental resistance $R=8ohms$

Resistor Feed $R_f=82ohms$

Supply Change $\triangle V=1$

Generally the equation for voltage rate of change is mathematically given by

$\frac{dV_0}{dV}=\frca{R}{R_1r_3}$

Therefore

$\triangle V_0=\triangle V*\frac{R}{R_fR}$

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$\triangle V_0=0.08V$

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Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

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so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

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Tr=<u>304,000</u>

900

=33.77

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