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ludmilkaskok [199]
4 years ago
14

(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific

weight of 66.4 lb/ft3. Determine the vertical force required to give the tank an upward acceleration of 10.7 ft/s2.
Engineering
1 answer:
lina2011 [118]4 years ago
5 0

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content =  66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

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Sedbober [7]

Answer:

The correct answer will be "400.4 N". The further explanation is given below.

Explanation:

The given values are:

Mass of truck,

m = 600 kg

g = 9.8 m/s²

On equating torques at the point O,

⇒  T\times Cos(10+5)\times (1.3+4)=mg\times Sin(5)\times 4

So that,

On putting the values, we get

⇒  T\times Cos(15^{\circ})\times 5.3=600\times 9.8\times Sin(5^{\circ})\times 4

⇒                             T=400.4 \ N

8 0
4 years ago
Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro
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Answer:

The answer is "0.0728"

Explanation:

Given value:

P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \  pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\

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if P flow is chocked

if P>P^{*} \to flow is not chocked

When  P= 10 psia < P^{*} \to not chocked

match number:

\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}

                       = \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}

M_0=7.625

p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}

  =0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}

\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\

\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\

                                    =\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\

R= gas constant=1716

m=PAV\\\\

    =0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}

5 0
3 years ago
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lesya [120]

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5 0
3 years ago
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Answer:

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