For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of alloy that is liquid.

Explanation:

The purpose of this program is to recognize the Youth's vital role in nation- building, promote consciousness among youth and develop their physical, moral, spiritual, intellectual and social well-being. It shall inculcate in the youth patriotism, nationalism, and advance their involvement in public and civic affairs.

Answer: true

Explanation:

Answer:

A diesel engine is a type of heat engine that uses the internal combustion process to convert the energy stored in the chemical bonds of the fuel into useful mechanical energy. ... First, the fuel reacts chemically (burns) and releases energy in the form of heat.

Answer:

Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.

Explanation:

Hi please follow me also I you can and thanks for the points. Have a good day.

Answer:

0.0432 m^3/s

Explanation:

Internal diameter of smaller pipes = 2.5 cm = 0.025 m

pipa wall thickness = 3 mm = 0.003 m

internal diameter of larger pipes = 8 cm = 0.8 m

velocity of region between smaller and larger pipe = 10 m/s

Calculate discharge in m^3/s

First we calculate the area of the smaller pipe

A = [tex]\pi Dt[/tex] = [tex]\pi ( 0.025 ) ( 0.003 )[/tex]  = 0.00023571 m^2

next we calculate area of fluid between the smaller pipes and larger pipe

A = [tex][\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})[/tex]

   = [tex][ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )][/tex]

   = [ 0.00502857 - 0.00070713 ]

   = 0.00432144 m^2

hence the discharge in m^3/s

Q = AV

   = 0.00432144 * 10

   = 0.0432 m^3/s

The question is incomplete. The complete question is --

An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :

1. the interference value, i

2. the radial pressure at the interference of the collar and the rod

3. the maximum effective stress in the collar

4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break

Solution:

Given :

Inner radius of the collar = 1 cm

So, inner diameter, [tex]$d_1$[/tex] = 2 cm

Outer radius of the collar = 12 cm

So, outer diameter, [tex]$d_2$[/tex] = 4 cm

The aluminium rod diameter, d = 1.01 cm

Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.

1. So, the interference , i =0

2. The radial pressure is also 0.

3. There will be no stress developed. So the maximum effective stress is 0

4. The collar will not break

Answer:

  il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

  il(t) = e^(-t/.01)

  il(t) = e^(-100t) . . . amperes

Answer:

It can be simulated with RTFO and PAV method

Explanation:

The aging which occurs for the binders is always tested using rolling thin film oven and also pressure aging vessels. The rolling thin film oven is a method used to calculate the short term aging that occurs in the asphalt cement when it mixes with the aggregates as well as in service. While the long term process of aging in asphalt cement is carried out by pressure aging vessel.

The short term process of aging can be simulated with the help of RTFO and the long term is simulated by the PAV method.

Answer:

True :)

Explanation:

If this is a true or false question.

The discoloration is when the color of a particular thing comes off or changes, fading from its original color. In other words, when the original color of anything changes or wanes off, then that is called discoloration. The statement is true.

This chemical shift occurs for a number of reasons, including its interaction with other elements nearby or the unfavorable effects of other elements it comes into touch with.

Water leaks may create cases of discoloration on walls, surfaces, even ceilings, and pipes. As a result, the surface progressively changes and becomes moist.

Discoloration puts the surrounding region in danger in addition to altering the part(s) or area's look and durability.

Since a change in that region or component is required, the raw material(s) used are wasted. The claim that discoloration results in the waste of raw materials are thus accurate.

Learn more about discoloration here:

brainly.com/question/6378613

#SPJ6

Answer:

1/2 semesters before completion

Explanation:

Program completion and review are both a necessary part and parcel of a successful completion of any program. The recommended period for applying for program completion and review is one to two semesters before the program is to be completed. This is mandatory, so that there is different time span for review and any other issues that are needed to be carried out. And then afterwards, finally the certificates are then issued to the qualified person.

Answer:

i'll help you but there is no question to answer??

Answer:

a. cooling of magma

Explanation:

they form when magma cools

Answer:

Earthquake's put vibration and seismic load on a building while wind puts a continuous force throughout the whole structure.

Explanation:

The most important difference between the design of a building that's resistent to earthquakes and resistant to high speed winds is that earthquake's put vibration and seismic load on a building while wind puts a continuous force throughout the whole structure.

Earthquake resistent building are equipped with a base that is isolated from the buildings with several layers of different material which absorb the vibration of the seismic waves and prevent them from reaching to the building.

Wind resistent buildings must have strong foundations and a sturdy structure/material that can withstand high pressure that the wind applies during events such as hurricanes.

I hope this answer helps.  

Answer:

115 Ib/ft^3

Explanation:

To determine the maximum dry density in Ib/ft3 we have to calculate :

Bulk unit weight ( yb ) ; W / v

Dry unit weight ( yd. ) :  yb / ( 1 + w )

For every set of data given

assuming  v = 1/30 ft^3

calculating for the 3 data set ( maximum dry density )

weight of soil (W) = 4.40

moisture content (%) (w) = 15.0 = 0.15

Bulk unit weight (yb) = 4.40 / (1/30)  = 132 Ib/ft^3

Dry unit weight ( yd. ) = 132 / ( 1 + 0.15 ) = 114.702 Ib/ft^3

therefore after calculations the maximum dry density in Ib/ft^3 ≈ 115 Ib/ft^3

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Answer:

c

Explanation:

the pluming usually is on the bottom floor plan

Answer:

Happy

Explanation:

Answer:

plz follow in titkok

Explanation:

Solution :

Given

The number of line that can be written by printer per page = 50 lines

The number of characters that can be written by printer per page = 80 characters.

Number of pages that can be written per minute = 6 pages

The speed for writing characters to the printer can be found by

= 50 x 80 x 6

= 400 characters per sec

In one second the printer has the writing speed of 400 character.

Now each of the character uses = 50 µsec of the CPU time for interrupt.

So in each second, the interrupt overhead = 20 msec

Now using the interrupt driven input/output,  980 msec time can be available for other work.

Thus, the interrupt overhead charges only 2 percent of the CPU, that will hardly affect the program to run.

Solution :

Given

[tex]$t_1=2+x_1$[/tex]

[tex]$t_2=1+x_2$[/tex]

Now,

[tex]$P(h<5)=1-P(h \geq5)$[/tex]

[tex]$0.4=1-P(h \geq5)$[/tex]

[tex]$0.6=P(h \geq5)$[/tex]

[tex]$0.6= e^{\frac{-x_1 5}{3600}}$[/tex]

Therefore,   [tex]$x_1=368 \ veh/h$[/tex]

                        [tex]$=\frac{368}{1000} = 0.368$[/tex]

Given,   [tex]$t_1=2+x_1$[/tex]

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, [tex]$t_2=t_1$[/tex]

∴  [tex]$t_2$[/tex] = 2.368 min

[tex]$t_2=1+x_2$[/tex]

[tex]$2.368=1+x_2$[/tex]

[tex]$x_2 = 1.368$[/tex]

[tex]$x_2 = 1.368 \times 1000$[/tex]

    = 1368 veh/h

Answer:

I think it's referred as the outline

Answer:

thanks! :D

Have a good day!

Answer:

Yay, :)

Thank you!!!

Answer:

S = 0.10253 kW/k

Explanation:

Given data:

Velocity of oxygen ( V ) = 70 m/s  

Diameter of pipe = 12-cm = 0.12 m

At entrance

pressure of oxygen ( p1 ) = 240 kPa

Temperature of oxygen ( T1 ) = 20°c  = 293 k

At exit

pressure of oxygen ( p2 ) = 200 kPa

temperature of oxygen ( T2 ) = 18°c = 291 k

First calculate specific volume of oxygen at inlet

V1 = [tex]\frac{RT1}{P1}[/tex]  -------- ( 1 )

  R = 0.2598 KJ/kgk ( property of oxygen )

  T1 = 293 k

  P1 = 240 kpa

substitute values into equation 1

V1 = 0.3172 m^3/kg

next we calculate the mass flow rate of Oxygen

m = [tex]\frac{A1V}{v1}[/tex] ----- ( 2 )

A1 ( area of pipe ) = 0.0113 m^2 ( calculated )

V = 70 m/s

V1 = 0.3172 m^3/kg

substitute value into equation 2

m ( mass flow rate of oxygen ) = 2.4936 kg/s

Finally calculate the rate at which entropy is generated in the pipe

S = [tex]m( C_{p} In\frac{T2}{T1} - RIn\frac{P2}{P1} )[/tex]   --------- ( 3 )

[tex]C_{p}[/tex] = 0.918 kj/kgk  ( property of oxygen )

T2 = 291 k

T1 = 293 k

P2 = 200 kPa

P1 = 240 kPa

substitute values into equation 3 above

S = 0.10253 kW/k

Solution :

Given

Volume, [tex]$V_1 = 1 \ m^3$[/tex]

Temperature, [tex]$T_1=20 \ ^\circ C[/tex]

[tex]$x_1=0.5$[/tex]

From the saturated water table, corresponding to [tex]$T_1=20 \ ^\circ C[/tex], we get the saturated liquid, vapor specific and the entropy.

[tex]$v_f=1.0010 \ m^3/kg$[/tex]

[tex]$v_g=57.791 \ m^3/kg$[/tex]

[tex]$s_f=0.2966 \ kJ/kg-K$[/tex]

[tex]$s_g=8.6672 \ kJ/kg-K$[/tex]

Now calculating the initial specific volume

[tex]$v_1=v_f+x_1 \cdot(v_g-v_f)$[/tex]

    [tex]$=1.0018+05 \cdot(57.791-1.0018)$[/tex]

    [tex]$= 29.8973 \ m^3/kg$[/tex]

Calculating the initial specific entropy:

[tex]$s_1=s_f+x+1 \cdot (s_g-s_f)$[/tex]

    [tex]$=0.2966+0.5 \cdot (8.6673 - 0.2966)$[/tex]

    [tex]$= 4.48 \ kJ/kg-K$[/tex]

So final volume of the vessel is two times bigger as the initial volume

[tex]$V_2=2 .V_1$[/tex]

    [tex]$= 2 \times 29.8973 = 59.8 \ m^3/kg$[/tex]

If we interpolate the values from tables between [tex]$v_g=57.791 \ m^3/kg$[/tex] and [tex]$v_g=61.293 \ m^3/kg$[/tex], we can get final temperature and specific entropy corresponding to value of [tex]$v_2$[/tex] :

Final temperature, [tex]$T_2= 19.6 ^\circ C$[/tex]

and [tex]$s_2 = 8.68 \ kJ/kg-K$[/tex]

Calculating change in entropy

[tex]$\Delta s = s_2-s_1$[/tex]

     [tex]$=8.68-4.48 = 4.2 \ kJ/kg-K$[/tex]

Answer:

500 feet, 300 feet

Explanation: