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Finger [1]
3 years ago
7

A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac

e. As it is being transported downwards in an elevator, someone accidentally presses the emergency stop button, such that the elevator comes to rest with a deceleration of magnitude 2g, where the gravitational acceleration g = 9.81 m/s2 . Calculate the pressure difference between the bottom and the surface of the tank.
Engineering
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a )  ------ ( 1 )

From the above data given will  be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1  becomes

Py =  -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

    = 3270 N/m^2

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vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Repo
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Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= density\times g\times \frac{h}{2}

By putting values, we get

= 1000\times 9.8\times \frac{12.2}{2}

= 1000\times 9.8\times 6.1

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hence,

The average force will be:

= Pressure\times Area

= 59780\times 3.6\times 12.2

= 2625537 \ N

Or,

= 2625 \ kN

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. A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The allowable shear stress for the material of the shaft is 42 MPa.
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3 years ago
The head difference between the inlet and outlet of a 1km long pipe discharging 0.1 m^3/s of water is 0.53 m. If the diameter is
Reptile [31]

Answer:

The correct option is 'e': f = 0.05.

Explanation:

The head loss as given by Darcy Weisbach Equation is as

h_{l}=\frac{flv^{2}}{2gD}

where

h_{l} is head loss in the pipe

'f' is the friction factor

'l' is the length of pile

'v' is the velocity of flow in pipe

'D' is diameter of pipe

From equation of contuinity we have v=\frac{Q}{A}

Thus using this in darcy's equation we get

h_{l}=\frac{flQ^{2}}{2gDA}

where

'Q' is discharge in the pipe

'A' is area of the pipe A=\frac{\piD^2}{4}

Applying the given values we get

h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}

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f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05

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