Determine the activation energy for diffusing copper into gold if you know that the diffusion coefficient is 3.98 x 10-13 m2/s at 980 degreeC and the diffusion coefficient is 3.55 x 10-16 m2/s at 650degreeC.

Answer:

c)Conduction

Explanation:

As we know that conduction heat transfer is take place due to movement of electron.When we fall on the campfire then the primary heat transfer mechanism is conduction because .

Convection heat transfer taker place due to motion of fluid.

Radiation heat transfer dominates at very high temperature but in campfire the temperature is not too high.

So option C is correct.

Answer:

The given statement is correct.

Explanation:

When we load a material axially stresses are developed in the material. these stresses arise internally to keep the object in equilibrium.

When a tensile load is applied to an material since the nature of the force is to cause elongation in the material, axial strain is developed in the material or we can say that any arbitrary point in the material undergoes a displacement in the direction of the applied load. The stresses that are developed in the material tend to tear the material apart.

Strength of a material by definition is it's ability to resist deformation, thus tensile strength can be defined as the resistance of the material to tensile strains which have the tendency to tear the material apart.

Answer:

Option c is True

Explanation:

a)- Machines are structures made of one or more movable parts. is false statement because Machines can be made without any movable part as in simple machines like inclined plains, wedges, screw etc.

b) Machines have a least one multi force member as part. is false statement since it is not a compulsory condition.

c) Machines transmit force or alter the effect. This is a true statement as machine are made in  the first place to transmit forces and alter the effect.  

d) Machines can be Contrained.

Answer:

Explanation:

Engineering Stress is defined as Load applied to the original cross-sectional area which we have taken in the start.

True stress is defined as the load divided by area of cross-section of specimen at that instant.

Engineering stress  and true stress can be expressed by relation

[tex]\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )[/tex]

Where

[tex]\sigma _T=True\ stress[/tex]

[tex]\sigma _E=Engineering\ stress[/tex]

[tex]\epsilon _E=Engineering\ Strain[/tex]

Answer:

Spring

Explanation:

We know that energy associated with motion is known as kinetic energy and the energy associated with position or elevation is known as potential energy.

A ball on floor which on the rest condition have zero potential energy but on the other hand a ball at a height h from the floor have mgh potential energy.

Spring is the mechanical component which stored potential energy.

Answer:

Mass of earth will be [tex]M=5.96\times 10^{24}kg[/tex]

Explanation:

We have given acceleration due to gravity [tex]g=9.81m/sec^2[/tex]

Radius of earth = 6370 km =[tex]6370\times 10^3m[/tex]

Gravitational constant [tex]G=6.67\times 10^{-11}Nm^2/kg^2[/tex]

We know that acceleration due to gravity is given by

[tex]g=\frac{GM}{R^2}[/tex], here G is gravitational constant, M is mass of earth and R is radius of earth

So [tex]9.81=\frac{6.67\times 10^{-11}\times M}{(6370\times 10^3)^2}[/tex]

[tex]M=5.96\times 10^{24}kg[/tex]

So mass of earth will be [tex]M=5.96\times 10^{24}kg[/tex]

Answer:

a)T=95.5414 N.m

b)d=21.35 mm

Explanation:

Given that

P=3.5 KW

Speed N=350 RPM

Yield strength= 100 MPa

So Shear strength = 0.5 x 100 =50 MPa

We know that

[tex]P=\dfrac{2\pi NT}{60}[/tex]

Where N is the speed ,T is the torque and P is the power.

Now by putting the values

[tex]P=\dfrac{2\pi NT}{60}[/tex]

[tex]3500=\dfrac{2\pi 350T}{60}[/tex]

T=95.5414 N.m

T=95,541.4 N.mm

We also know that

[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]

d is the diameter of shaft

[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]

[tex]50=\dfrac{16\times 95541.4}{\pi d^3}[/tex]

d=21.35 mm

Torque,T=95.5414 N.m

Diameter ,d=21.35 mm

#WeDemandDemocraticEritreaNow

Answer:

Silica is the principal component in glass.

Explanation:

Step1

Glass is the amorphous solid and transparent. Glass products have many of the shapes and design that are available in market.

Step2

Natural quartz is the primary source of glass in sand. Silica is the principal component in approximately all glass. Lime stone, soda ash and aluminum oxide are added in the galas for desired properties depending upon application. So, silica is the principal component in glass.

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of [tex]300\times 100 mm^2[/tex]

Force(F)=8 MN

Therefore stress due to this Force([tex]\sigma [/tex])

[tex]\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}[/tex]

[tex]\sigma =266.66 \times 10^{6} Pa[/tex]

[tex]\sigma =266.66 MPa[/tex]

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

Answer:

Shearing strain will be 0.1039 radian

Explanation:

We have given change in length [tex]\Delta L=0.12inch[/tex]

Length of the pad L = 1.15 inch

We have to find the shearing strain

Shearing strain is given by

[tex]\alpha =tan^{-1}\frac{\Delta L}{L}=tan^{-1}\frac{0.12}{1.15}=5.9571^{\circ}[/tex]

Shearing strain is always in radian so we have to change angle in radian

So [tex]5.9571\times \frac{\pi }{180}=0.1039radian[/tex]

Calculating the force of attraction between a cation and an anion using Coulomb's Law involves converting the given distance into meters, using the elementary charge for valences, and then applying the formula.

The question is about calculating the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, separated by a distance of 2.9 nm. To find this, we use Coulomb's Law, which is given by the equation F = k * |q1*q2| / r2, where F is the force between the charges, k is Coulomb's constant (8.9875517873681764 × 109 N m2 C-2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges. However, given the units of distance as nanometers (nm) and charge in terms of electron charge, we first convert the distance to meters (1 nm = 1 × 10-9 meters) and use the elementary charge (1.6022 x 10-19 C) for the charges. With a +2 valence for the cation and a -1 valence for the anion, the charges become +2e and -1e, respectively. The distance r is 2.9 nm or 2.9 × 10-9 meters. Plugging these values into Coulomb's formula, the attraction force can be calculated, taking into account that the force will be negative, signifying attraction.

Answer:

1.Kinetic energy  2.Enthalpy

Explanation:

We know that diffuser is a device which convert kinetic  energy of fluids into pressure energy of fluids,on the other hand nozzle is used to convert the pressure energy of fluids into kinetic energy of fluids. If we want exit velocity of fluid is too high then we use nozzle and if we wand exit velocity of fluid is low then we use diffuser.

We know that enthalpy is also related with pressure so we can  say that ,diffuser covert kinetic energy into enthalpy.

Enthalpy h= u + p.v

Where u is the internal energy ,p is the pressure and v is volume.

Answer:

Time needed to empty the pool is 401.35 seconds.

Explanation:

The exit velocity of the water from the orifice is obtained from the Torricelli's law as

[tex]V_{exit}=\sqrt{2gh}[/tex]

where

'h' is the head under which the flow of water occurs

Thus the theoretical discharge through the orifice equals

[tex]Q_{th}=A_{orifice}\times \sqrt{2gh}[/tex]

Now we know that

[tex]C_{d}=\frac{Q_{act}}{Q_{th}}[/tex]

Thus using this relation we obtain

[tex]Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}[/tex]

Now we know by definition of discharge

[tex]Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}[/tex]

Using the above relations we obtain

[tex]Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\[/tex]

The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0

Solving for 't' we get

[tex]\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.[/tex]

Answer:

d. to supply heat to a warmer space

Explanation:

Let's analyze the options

(a) to remove heat from a colder space is the purpose of a refrigerator. Note that heat pump also do this  

(b) to remove heat from a warmer space. This option has no sense, both refrigerators and heat pumps supply heat to a warmer space

(c) to supply heat to a colder space. This option has no sense, both refrigerators and heat pumps remove heat from a colder space

(d) to supply heat to a warmer space. This is the purpose of a heat pump

Answer:

The lamp will remain off.

Explanation:

I assume this is a simple series circuit with a battery, a switch, an incandescent lamp and a capacitor.

Since the circuit is of continuous current (because it uses a battery), after the switch has remained closed for a long time the capacitor will be fully charged. Being fully charged, no continuous current will flow through the capacitor, and since it is in series with the rest of the circuit there will be no current anywhere, so the lamp will remain off.

Answer with Explanation:

The relation between power and energy is

[tex]Energy=Power\times Time[/tex]

Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals

[tex]E=1200\times 10^{6}\times 3600\times 24\times 365=3.784\times 10^{16}[/tex]

Now from the energy mass equivalence we have

[tex]E=mass\times c^2[/tex]

where

'c' is the speed of light in free space

Thus equating both the above values we get

[tex]3.784\times 10^{16}=mass\times (3\times 10^{8})^{2}\\\\\therefore mass=\frac{3.784\times 10^{16}}{9\times 10^{16}}=0.42kg[/tex]

Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is

[tex]mass_{req}=\frac{mass}{\eta }=\frac{0.43}{0.34}=1.235[/tex]

Thus 1.235 kg of nuclear fuel is reuired for operation.

Answer:

0.948 Btu

Explanation:

1 Btu = 1055 J so [tex]\frac{1000}{1055}[/tex] = 0.948 Btu

1 BTU = 1055.06 joules **

(1000 J) x (1 BTU / 1055.06 J) = 0.9478 BTU

** Note:  You'll have to accept my conversion factor, since I went and looked it up and you didn't.

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be [tex]Q\frac{gallons}{minute}[/tex]

Now replacing the gallon and the minute by the above values we get

[tex]Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}[/tex]

Thus [tex]Q'=0.631Q\frac{liters}{second}[/tex]

Now since we know that 1 liter = [tex]0.0353ft^{3}[/tex]

Using this in above relation we get

[tex]Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q[/tex]

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

Answer:

21.456 kJ/h

Explanation:

See the figure attached. In this case  

[tex]W_{cycle} = 4 kW [/tex]

[tex]Q_{out} = \text{heating effect}[/tex]

Coefficient of performance in heat pump is defined by

[tex]COP = \frac{Q_{out}}{W_{cycle}} [/tex]

[tex]Q_{out} =COP*W_{cycle} [/tex]

[tex]Q_{out} =1.49*4 \, W [/tex]

[tex]Q_{out} = 5.96 \, W [/tex]

Now it is necessary to change units, remember that Watt (W) is defined as J/s

[tex]Q_{out} = 5.96 \frac{J}{s} \frac{3600s}{1 h} \frac{1 kJ}{1000 J}[/tex]

[tex]Q_{out} = 21.456 \frac{kJ}{h} [/tex]

Answer:

Net discharge per hour will be 3.5325 [tex]m^3/hr[/tex]

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius [tex]r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m[/tex]

We know that area is given by

[tex]A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2[/tex]

We know that discharge is given by [tex]Q=AV[/tex], here A is area and V is velocity

So [tex]Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec[/tex]

So net discharge in 1 hour = [tex]981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour[/tex]

Explanation:

Step1

Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.  

Step2

Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product.  So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.

Answer:

a) 333,33 b)291,67

Explanation:

you have both equations :

C=50+0.5V and V=2500-10C

and you want to know the variable V so, you can calculate V in function of C, and you have already clear the variable "C", then you replace for (a):

for b) they tell you that you increase your constant of the equiation C=50+0.5V (remember that the constant is the one alone, wihout any variable, in this case "50") increase in 25 so, your equiation for this point of C is C=75+0.5V, then you do the same:

It's a system of two equations and two variables, wich gives you a compatible define system, it gives you only one solution.

I hope it helps you.

To determine the volume of traffic across the bridge, we can solve an equation relating cost and traffic volume. The volume of traffic across the bridge is approximately 333 vehicles per day. If a toll of 25 cents is added, the volume of traffic across the bridge is approximately 292 vehicles per day.

To determine the volume of traffic across the bridge, we need to solve the equation for V. The equation relating cost and volume of traffic is given by V = 2500 - 10C. Substituting the cost function C = 50 + 0.5V into the equation, we get V = 2500 - 10(50 + 0.5V). Simplifying the equation, we find V = 2500 - 500 - 5V. Combining like terms, we get 6V = 2000, which yields V = 333.33. Therefore, the volume of traffic across the bridge is approximately 333 vehicles per day.

To find the volume across the bridge if a toll of 25 cents is added, we can modify the cost function to be C = 50 + 0.5V + 25 cents. We again substitute this into the equation V = 2500 - 10C and solve for V. Substituting in the new cost function, we get V = 2500 - 10(50 + 0.5V + 25). Simplifying the equation, we find V = 2500 - 10(75 + 0.5V). Continuing to simplify, we have V = 2500 - 750 - 5V. Combining like terms, we get 6V = 1750, which yields V = 291.67. Therefore, the volume across the bridge with the toll of 25 cents is approximately 292 vehicles per day.

Answer:

308 acre-ft of water

Explanation:

Given:

Area of the watershed = 22 square miles

Depth of Rainfall = 0.75 in =[tex]\frac{\textup{0.75}}{\textup{12}}[/tex]=0.0625 ft

Percentage rainfall falling in reservoir as runoff = 35%

Now,

1 square mile = 640 acre

Thus,

22 square miles = 22 × 640 = 14,080 acres

Thus,

The total volume of rainfall = Area of watershed × Depth of the rainfall

or

The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft

also,

only 35% of the total rainfall is contributing as runoff

thus,

Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water

Answer:

The collision force is 81987.577 lbf

Explanation:

Apply Newton’s second law of motion for required collision force exerted on the goose.

Given:  

Mass of goose is 12 lb.

Time of collision is [tex]10^{-3}[/tex] s.

Taking off speed is 150 miles/h.

Calculation:  

Step1

Convert take off speed in ft/s as follows:

[tex]v=(150mi/h)(\frac{\frac{5280}{3600}ft/s}{1mi/h})[/tex]

v = 220 ft/s

Step2  

Collision force is calculated as follows:

[tex]F=\frac{mv}{t}[/tex]

[tex]F=\frac{(\frac{12}{32.2})220}{10^{-3}}[/tex]

F = 81987.577 lbf.

Thus, the collision force is 81987.577 lbf.  

Answer and Explanation:

AIR PREHEAT ER :

Answer:

P=25000lbf

Explanation:

For this problem we will use the equation that relates, the effort, the area and the force for an element under normal stress.

σ=P/A

σ=stress=20kSI=20 000 lbf/in ^2

P=force

A=area

solving for P

P=Aσ

P=(20 000 lbf/in ^2)(1.25in^2)

P=25000lbf

Answer:

The maximum magnitude P of the loads that can be applied to the truss = 25,000 Pounds or 111,205.5 Newtons.

Explanation:

In order to calculate the maximum load P, we will make use of the formula: Maximum average stress (20 ksi) = maximum load P ÷ cross-sectional area (1.25 in²)

Make P (the maximum load) the subject of the formula: P = 20 ksi × 1.25 in².

Before moving further, we have to convert the average normal stress (in ksi) to an appropriate unit: The average normal stress = 20 ksi = 20 kip per square inch (kip/in²)

But 1 kip = 1000 Pounds (i.e., 1000 lb)

Therefore, 20 ksi = 20,000 Pounds/in².

Therefore, P (maximum load) = 20,000 pounds/in² × 1.25 in² = 25,000 Pounds = 111,205.5 Newtons (because 1 Pound = 4.44822 Newtons).

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by  [tex]\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}[/tex]

So availability of system [tex]=\frac{900}{900+100}=\frac{900}{1000}=0.9[/tex]

So the availability of system will be 0.9

Answer:

25.4 kW

Explanation:

There are 15 people doing exercises, each will dissipate 740 W of heat, so they will disspiate a total of p = 15*740 = 11100 W = 11.1 kW

There are 7 treadmills, each has a 2.5 hp motor (1.86 kW) running at a load factor of 0.7 with an efficiency of 0.77. So their total power would be p = 7*1.86*0.77/0.7 =14.3 kW

So the total heat dissipated would be 11.1 + 14.3 = 25.4 kW.

The rate of heat gain from people and equipment at peak load is approximately 14.955 kW, including treadmill and people's heat dissipation.

To calculate the rate of heat gain of the exercise room from people and equipment at peak load conditions, we need to consider the heat dissipated by both the exercising equipment and the people in the room.

1. Heat dissipation from exercising equipment:

  - For the treadmills: Each treadmill has a 2.5-hp motor operating at 0.7 load factor and 0.77 efficiency. So, the power consumed by each treadmill is [tex]\( P_{\text{treadmill}} = 2.5 \, \text{hp} \times 0.7 \times 0.77 = 1.925 \, \text{kW} \).[/tex]

  - For the weight-lifting machines: Since they have no motors, they do not contribute to heat dissipation.

2. Heat dissipation from people:

  - There are two people doing light exercises. Assuming each person dissipates heat at a rate of 740 W, the total heat dissipated by people is [tex]\( P_{\text{people}} = 2 \times 740 \, \text{W} = 1480 \, \text{W} \).[/tex]

Now, we can calculate the total rate of heat gain of the exercise room:

[tex]\[ \text{Total heat gain} = P_{\text{treadmill}} \times \text{number of treadmills} + P_{\text{people}} \][/tex]

Given there are 7 treadmills, we have:

[tex]\[ \text{Total heat gain} = 1.925 \, \text{kW} \times 7 + 1480 \, \text{W} \]\[ \text{Total heat gain} = 13.475 \, \text{kW} + 1480 \, \text{W} \]\[ \text{Total heat gain} \approx 13.475 \, \text{kW} + 1.48 \, \text{kW} \]\[ \text{Total heat gain} \approx 14.955 \, \text{kW} \][/tex]

So, the rate of heat gain of the exercise room from people and equipment at peak load conditions is approximately 14.955 kW.

Answer:

1) Height of oil in right limb = 67.31 cm

2) Height of water in the right limb = 16.83 cm

Explanation:

The U-tube manometer is shown in the attached figure

Foe equilibrium The pressure at the bottom of the U tube should be same

Let the height of the water in the left limb of the manometer be [tex]h_L[/tex]

Thus the pressure at the bottom is found using the equation of pressure statics as

[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{L}.............(i)[/tex]

Similarly for the liquid in the right limb the pressure at the bottom is the sum of the oil column and the water column

Thus we can write

[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}...........(ii)[/tex]

Equating the equations 'i' and 'ii' we get

[tex]P_{atm}+\rho _{water}\times g\times h_{L}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}\\\\\rho _{water}\times g\times 0.7=\rho _{oil}\times 4h\times g+\rho _{water}\times h\times g\\\\\therefore h_{}=\frac{0.7\times \rho _{water}}{4\times \rho _{oil}+\rho _{water}}\\\\h_{}=\frac{0.7\times 1000}{4\times 790+1000}=16.83cm[/tex]

Thus the height of oil is [tex]4\times 16.83=67.31cm[/tex] andthe height of water in the right limb is 16.83 cm.

Answer:

a)Δs = 834 mm

b)V=1122 mm/s

[tex]a=450\ mm/s^2[/tex]

Explanation:

Given that

[tex]s = 15t^3 - 3t\ mm[/tex]

a)

When t= 2 s

[tex]s = 15t^3 - 3t\ mm[/tex]

[tex]s = 15\times 2^3 - 3\times 2\ mm[/tex]

s= 114 mm

At t= 4 s

[tex]s = 15t^3 - 3t\ mm[/tex]

[tex]s = 15\times 4^3- 3\times 4\ mm[/tex]

s= 948 mm

So the displacement between 2 s to 4 s

Δs = 948 - 114 mm

Δs = 834 mm

b)

We know that velocity V

[tex]V=\dfrac{ds}{dt}[/tex]

[tex]\dfrac{ds}{dt}=45t^2-3[/tex]

At t=  5 s

[tex]V=45t^2-3[/tex]

[tex]V=45\times 5^2-3[/tex]

V=1122 mm/s

We know that acceleration a

[tex]a=\dfrac{d^2s}{dt^2}[/tex]

[tex]\dfrac{d^2s}{dt^2}=90t[/tex]

a= 90 t

a = 90 x 5

[tex]a=450\ mm/s^2[/tex]