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katrin2010 [14]
3 years ago
14

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

he water rise? if the plates are placed in mercury at 20 degree celcius, how far will the column of mercury be depressed?
Engineering
1 answer:
Ulleksa [173]3 years ago
4 0

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

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Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}

1x0.723x(T_{eq} -350)=3x0.780x(450-T_{eq} ) ⇒T_{eq} = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

V_{a,i}  = \frac{mRT_{a,i} }{P_{a,i} }=  \frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar} = 0.201m^{3}

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }

P_{eq}= 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

                             (0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0
3 years ago
Isormophous phase diagram
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Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

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Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed
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Explanation:

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4 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

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3 years ago
What is the mechanical advantage of a pulley with 3 support ropes?
snow_tiger [21]

Answer:

The mechanical advantage is 3 to 1

Explanation:

A frictionless pulley with three support ropes carries equal tension on each of the ropes thus;

Tension in each pulley rope = T

Total tension in the 3 ropes = 3 × T = 3·T

Direction of the tension forces on each rope = Unidirectional

Total force provided by the 3 ropes = 3·T

Therefore, a force, T, applied at the end of the rope will result in a lifting force of 3·T

Hence, the mechanical advantage = 3·T to T which is presented as follows;

Mechanical \ advantage = \dfrac{3 \cdot T}{T}  = \dfrac{3}{1}

The mechanical advantage = 3 to 1.

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3 years ago
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