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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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An engine has a piston with a surface area of 17.31 in2 and can travel 3.44 inches. What is the potential change in volume, disp
Katena32 [7]

Answer:

$$\begin{align*}

P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$

Explanation:

\eqalign{

 P(Y-X=m\mid Y\gt X)

   &=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr

   &=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr

   &=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr

}

P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$

5 0
3 years ago
-0-1"<br> -0<br> -20<br> -15<br> -10<br> 0<br> -5
kari74 [83]

Answer:

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Explanation:

what is that

3 0
3 years ago
ABS system is necessary?
Monica [59]

Explanation:

I think it helps you

I don't know the answer sorry

3 0
3 years ago
Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th
Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

PV=mRT

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

T_1 = 38\°C

T_2 = 14\°C

\eta = 97\%

\dot{v} = 510m^3/kg

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

PV = mRT

\dot{m} = \frac{PV}{RT}

\dot{m} = \frac{0.6626*10^{5}*510}{287*311}

\dot{m} = 37.85kg/min

Therfore the mass flow rate at which water condenses, then

\eta = \frac{\dot{m_v}}{\dot{m}}

Re-arrange to find \dot{m_v}

\dot{m_v} = \eta*\dot{m}

\dot{m_v} = 0.97*37.85

\dot{m_v} = 36.72 kg/min

PART B) Enthalpy is given by definition as,

H= H_a +H_v

Where,

H_a= Enthalpy of dry air

H_v= Enthalpy of water vapor

Replacing with our values we have that

H=m*0.0291(38-25)+2500m_v

H = 37.85*0.0291(38-25)-2500*36.72

H = 91814.318kJ/min

In the conversion system 1 ton is equal to 210kJ / min

H = 91814.318kJ/min(\frac{1ton}{210kJ/min})

H = 437.2tons

The cooling requeriment in tons of cooling is 437.2.

3 0
3 years ago
A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro
Zarrin [17]

Answer:

  83.6%

Explanation:

<h3>(a)</h3>

On its current schedule, the plant can theoretically produce ...

  (30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

<h3>(b)</h3>

On the proposed schedule, this production becomes ...

  (30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

  = (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

  (2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

3 0
2 years ago
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