____________ is the range of all the colors created by different amounts of light.

Calculate the electroosmotic velocity of an aqueous solution through a glass capillary 5 cm long with a 0.5 mm internal diameter

natita [175]

Answer:

Electroosmotic velocity will be equal to $1.6\times 10^{-4}m/sec$

Explanation:

We have given applied voltage v = 100 volt

Length of capillary L = 5 mm = 0.005 m

Zeta potential of the capillary surface $\xi =80mV=0.08volt$

Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as $\epsilon =10$

Viscosity of glass is $\eta =10^8$

Electroosmotic velocity is given as $v_{eo}=\frac{\epsilon \xi }{\eta }\times \frac{v}{L}$

$v_{eo}=\frac{10\times 0.08 }{10^8 }\times \frac{100}{0.005}=1.6\times 10^{-4}m/sec$

So Electroosmotic velocity will be equal to $1.6\times 10^{-4}m/sec$

8 0

4 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

a)

$-V_{2} =2V_{1}$ ( double)

b)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

b)

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago

Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give

Lelechka [254]

Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path

Explanation:

The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction

5 0

4 years ago

Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1

KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2 and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x y

-2 -4

-1 -3

0 -2

1 -1

2 0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

5 0

4 years ago

Read 2 more answers

Design a fundamental mode asynchronous finite state machine that accepts input pair, A and B. The AB input sequence 00, 01,11, 1

dybincka [34]

Answer:

See explaination

Explanation:

A Finite state machines can be synchronous or asynchronous. The operation of asynchronous state machines does not require a clock signal. An Asynchronous state machine is classified basically on their operating mode, such as the fundamental mode, pulse mode or burst mode. An asynchronous state machine can have stable and transient states.

Please kindly refer to attachment for a step by step solution.

5 0

3 years ago