What is Potential flow?

Answer:

(a)Volume in liters=5.3 liters.

(b)Volume in liters/minute=31.8 liters/minute.

Explanation:

Given:  

   Diameter of cylinder ,D=150 mm

                         Stroke,L=300 mm

                         Time ,t=10 sec

we know that swept volume of cylinder

          [tex]V_{s}=\dfrac{\pi }{4}\times D^2\times L[/tex]

So [tex]V_{s}=\dfrac{\pi }{4}\times 0.15^2\times 0.3 m^{3}[/tex]

     [tex]V_{s}=0.0053 m^3[/tex]

(a) Volume in liters =5.3 liters         ( 1[tex]m^3[/tex]=1000 liters)

(b) When we divide swept volume  by time(in minute) we will get liters/minute.

 We know that 1 minute=60 sec

⇒10 sec=[tex]\frac{10}{60}[/tex] minute

So volume displace in liters/minute=31.8 liters/minute.

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1-[tex]\frac{T2}{T1}[/tex]

=1-[tex]\frac{383}{873}[/tex]

=1-0.43871

=.56

=56%

Answer: True

Explanation:Typical metals have a property of ductility and malleability that is  metals can be drawn into wires or any other shape by beating or stretching the metal by putting the tensile strength or shear strength that pulls them apart . But while compression the metals are squeezed together which affects the hardness of a metal and they are not able to bear the compression force well and thus cannot show elastic-plastic behavior while compression .Therefore the statement given is true typical metals show elastic-plastic behavior in tension and shear but not in compression.

Answer:

b). False

Explanation:

Lumped body analysis :

Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.

                      Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.  

                      The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the  conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.

In general it is assume that for a lumped body analysis, Biot number [tex]\leq[/tex] 0.1

Therefore, the smaller the Biot number, the more exact is the lumped system analysis.

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)=[tex]\frac{Work produced}{Heat supplied}[/tex]

and

Heat Supplied [tex]{\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )[/tex]

[tex]{Q_s}=3.3+4.4=7.7KW[/tex]

η=[tex]\frac{3.3}{7.7}[/tex]

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1-[tex]\frac{T_ {sink}}{T_ {source}}[/tex]

η=1-[tex]\frac{273}{1023}[/tex]

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

Answer:

Rankine cycle less efficient as compare to Carnot cycle operating betwwen same temperature limit.

Explanation:

We know that Carnot's cycle is an ideal cycle for all heat engine which operating between same temperature.It is a reversible cycle which have all process reversible that is why it have maximum efficiency.

On the other hand Rankine cycle is a practical working cycle so it is impossible to make all process reversible .In practical there will be always loss due to this any process can not make 100 % reversible.That is why Rankine cycle have low efficiency as compare to Carnot cycle operating between same temperature limits.

Answer: The two different types of pumps in compressors are:

 1) Centrifugal Pumps

 2) Reciprocating Pumps

Explanation:

Compressor is defined as a mechanical device which increases the gas pressure by reducing its volume.The main action of a pump is to transport liquids and pressurize and it coverts rotational energy. As, centrifugal pumps are the most common pumps used for transfer of fluids and it works on simple mechanism.

Reciprocating Pumps is used for low volumes of flow at a high pressure. It is the positive displacement pump. As, it works on the principle of pushing of liquid which executes a reciprocating motion in a closed cylinder.  

Answer:

Centrifugal Pumps and Positive Displacement Pumps

Explanation:

The two of pumps listed above is based on the mode of operation. The centrifugal pumps works by increasing the velocity of the liquid through the machine while the displacement pumps works by alternating, filling a cavity and then displace some amount of liquid.

Answer:

    [tex]\mu_{min}[/tex]=[tex]26.38^{\circ}[/tex]

   [tex]\mu_{max}[/tex]=[tex]71.79^{\circ}[/tex]    

Explanation:

[tex]r_{1}[/tex]=7 in, [tex]r_{2}[/tex]=3 in,  [tex]r_{3}[/tex]=9in

       ,[tex]r_{4}[/tex]=8 in

  Transmission angle (μ ):

                   It is the acute angle between coupler and the output (follower) link.

Here we consider link [tex]r_{1}[/tex] as fixed link ,[tex]r_{2}[/tex] as input link ,link [tex]r_{3}[/tex] as coupler and link  [tex]r_{4}[/tex] as output link.

As we know that

[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]

[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]

When link [tex]r_{2}[/tex] will be horizontal in left side direction then transmission angle will be minimum and when link [tex]r_{2}[/tex] will be horizontal in right side direction then transmission angle will be maximum.

Now by putting the values we will find

[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]

[tex]\cos\mu_{max}=0.3125[/tex]

[tex]\mu_{max}=71.79^\circ[/tex]

[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]

[tex]\cos\mu_{min}=0.8958[/tex]

[tex]\mu_{min}=26.38^\circ[/tex]

Hence, The minimum and maximum angle of transmission angle is 26.38° and 71.79° respectively.

Answer Explanation :

ACCELEROMETERS:  Accelerometers are used for sensing the vibration. basically accelerometers are used for measuring high as well as low frequency. when a transducer is used in addition with another device to measure vibration is called pickups. Seismic instruments are commonly used  vibration pickups

PROXMITY PROBE: Proximity probe is used for the measurements of vibration the vibration sensitivity is highest around 8 to 16 hertz

Answer:

(i) 50 mm

(ii) 874.62 m N/m

(iii) 3116.45 m N/m

Explanation:

Given data

hexagonal head bolt = M14 x 2

steel plate = 15 mm

Young's modulus = 207 Gpa

Solution

1st part

dia of bolt (D) = 14 mm and height (H) = 12.8 mm from table

we know grip length = thickness of this plate i.e 30 mm , i.e (15mm+15mm)

so hexagonal bolt length = grip length + H

hexagonal bolt length = 30 mm + 12.8 mm = 42.80 mm i.e = 45 mm (round off)

2nd part

bolt stiffness =  [tex]\frac{Ad*At*young modulus}{Ad*Lt*At*Ld}[/tex]

here Lt is length of thread = 2d +6mm

d is 14 so length of thread = 2*14 +6 = 34 mm

At from table 8-L i.e. = 115 mm2

Ad area of without thread part = [tex]\pi /4[/tex]×[tex]d^{2}[/tex]

Ad= [tex]\pi /4[/tex]×[tex]14^{2}[/tex] = 153.94 mm2

Ld is length of bolt without thread = length of bolt - Lt = 45 -34 = 11 mm

last Lt thread part length = length of bolt - Ld = 30-11 = 19 mm

put all these value in bolt stiffness i.e.

bolt stiffness =  [tex]\frac{153.94*115*207}{153.94*19+115*11}[/tex] = 874.62

3rd part

stiffness of member =  [tex]\frac{0.5774 \pi  Ed}{2 ln (5\frac{0.5774L +0.5d}{0.5774L +2.5d})}[/tex]

here l is 30 and d is 14

so

stiffness of member =  [tex]\frac{0.5774 \pi  (207) 14}{2 ln (5\frac{0.5774(30) +0.5(14)}{0.5774(30) +2.5(14)})}[/tex]

stiffness of member =  3116.45 m N/m

Answer:

299.36 kPa

Explanation:

given mass of steam =2 kg

initial pressure that is [tex]P_1=500kPa[/tex]

initial temperature that is [tex]T_1=350^{\circ} C=350+273=623 K[/tex]

final temperature that is [tex]T_2=100^{\circ} C=100+273=373 K[/tex]

it is a rigid tank so volume is constant

for constant volume process [tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

[tex]P_2=\frac{T_2}{T_1}\times P_1=\frac{373}{623}\times 500=299.36 kPa[/tex]

so the final pressure will be 299.36 kPa

Answer:

b) False

Explanation:

Work done by a system is not a property because it doesn't define the system's state. Work is mechanical energy exchanged across the system's boundaries.

Answer:

a).Final temperature, [tex]T_{2}[/tex] = 180°C

b).Initial Volume, [tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]

   Final Volume, [tex]V_{2}[/tex] = 0.33012 [tex]m^{3}[/tex]

c). Initial enthalpy,[tex]H_{1}[/tex] =9129.68 kJ

   Final enthalpy, [tex]H_{1}[/tex] =6234.76 kJ

Explanation:

Given :

Total mass, m= 2.8 kg

Initial temperature, [tex]t_{i}[/tex] = 400°C

Initial pressure, [tex]p_{i}[/tex] = 1.2 MPa

Therefore from steam table at 400°C, we can find--

[tex]h_{1}[/tex] = 3260.6 kJ/kg

[tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg

Now it is mentioned that 28% of the mass is condensed into liquid.

So, mass of liquid, [tex]m_{l}[/tex] = 0.28 of m

                                                        = 0.28 m

     mass of vapour, [tex]m_{v}[/tex] = 0.72 m

∴ Dryness fraction, x = [tex]\frac{m_{v}}{m_{l}+m_{v}}[/tex]

                                  = [tex]\frac{0.72 m}{0.28 m+0.72 m}[/tex]

                                  = 0.72

a). The final temperature can be evaluated from the steam table at 1.2 MPa,

     [tex]h_{2}[/tex] = 2226.7 kJ/kg

     [tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg

    Final temperature, [tex]T_{2}[/tex] = 180°C

b). We know [tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg

    ∴ Initial Volume, [tex]V_{1}[/tex] = [tex]v_{1}[/tex] x m

                                [tex]V_{1}[/tex] = 0.25479 x 2.8

                                 [tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]

   We know,[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg

       ∴ Final Volume, [tex]V_{2}[/tex] = [tex]v_{2}[/tex] x m

                                    [tex]V_{2}[/tex] = 0.1179 x 2.8

                                   [tex]V_{1}[/tex] = 0.33012 [tex]m^{3}[/tex]

c). We know,

[tex]h_{1}[/tex] = 3260.6 kJ/kg

∴ Initial enthalpy,[tex]H_{1}[/tex] = [tex]h_{1}[/tex] x m

                                                    = 3260.6 x 2.8

                                                     = 9129.68 kJ

[tex]h_{2}[/tex] = 2226.7 kJ/kg

∴ Final enthalpy, [tex]H_{1}[/tex] = [tex]h_{2}[/tex] x m

                                                     = 2226.7 x 2.8

                                                     = 6234.76 kJ

Answer:

[tex]\omega_y,\omega_x,\omega_Z[/tex]  all are zero.

Explanation:

We know that if flow is possible then it will satisfy the below equation

[tex]\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0[/tex]

Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.

And velocity potential function [tex]\phi[/tex] given as follows

 [tex]u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}[/tex]

Rotationality of fluid is given by [tex]\omega[/tex]

[tex]\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z[/tex]

[tex]\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x[/tex]

[tex]\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y[/tex]

So now putting value in the above equations ,we will find

[tex]\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},[/tex]

[tex]\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}[/tex]

So [tex]\omega_y[/tex]=0

Like this all [tex]\omega_y,\omega_x,\omega_Z[/tex] all are zero.

That is why  velocity potential flow is irroational flow.

Answer Explanation:

the efficiency of the the engine is given by=1-[tex]\frac{T_2}{T_1}[/tex]

where T₂= lower temperature

           T₁= Higher temperature

we have given efficiency =70%

lower temperature T₂=27°C=273+27=300K

higher temperature T₁=627°C=273+627=900K

efficiency=1-[tex]\frac{T_2}{T_1}[/tex]

                =1-[tex]\frac{300}{900}[/tex]

                 =1-0.3333

                 =0.6666

                 =66%

66% is less than 70% so so inventor claim is wrong

Answer:

 Different types of equipment are required for proper conditioning of air because every air conditional space faces some geometrical and environmental issues or problems. There are some different types of equipment used for conditioning of air that are air system, water system and air-water system. In many cases the air conditioning of the system varies with size of the equipment.  

Answer Explanation:

DEFECT IN CASTING : Defects are undisireable its not matter which type of defect we want no defect in any process casting defect are observed in the material manufacturing in metal casting process.

THESE DEFECTS ARE CLASSIFIED INTO DIFFERENT CLASS :

NDT (NON DISTRUCTIVE TESTING): It is a inspection of material without distroy the material

there are different types of NDT

Answer:

Cold dies, weak metal temperature, dusty metal, lack of ventilation, and too much lubricant can cause casting defects.

Explanation:

Cold dies, weak metal temperature, dusty metal, lack of ventilation, and too much lubricant can cause casting defects. Gas permeability, shrinkage porosity, warm tears, and stream stains are other potential defects. Caused by poor gating, sharp edges or heavy lubricant, flow traces are mark left on the casting surface.

fiver techniques used to identify defects in material are shown below  

1) Liquid Penetrant Testing – Liquid penetrant screening is among the simplest approaches used to identify component defects.

2) Electromagnetic Testing – Electromagnetic testing comprises Eddy Current Testing, Rotating Current Field Quantification and Wireless Field Testing.  Above mentioned techniques can identify both surface &underground flaws.

3) Magnetic Particle Testing – Magnetic particle screening is commonly used for the ferromagnetic materials to detect ground and near-surface defects.

4) Ultrasonic testing – Ultrasonic testing makes it possible to identify large and extremely small surface defects

5) Thermal Infrared Testing –  Infrared thermography testing is used to measures and maps thermal variations on a substance's surface through thermal imaging equipment.

Answer:

a)[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]

b)[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]

c)[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]

Explanation:

a)

Adiabatic turbine

Adiabatic turbine means turbine can not reject or take heat from surrounding.

Isentropic efficiency of turbine can be define as the ratio of actual work out put to the Ideal or isentropic work out put.Ideal means when turbine will give maximum work and there is no any friction losses we can say when process is isentropic.

Isentropic efficiency of turbine=(Actual work output)/(Ideal work output)

[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]

b)

Adiabatic compressor

Isentropic efficiency of compressor can be define as the ratio of ideal or isentropic work in put to the actual work  in put.

Isentropic efficiency of turbine=(Ideal work input)/(actual work input)

[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]

c)Adiabatic nozzle

We know that nozzle is device which used to accelerate the fluid.

Basically it covert pressure energy to kinetic energy.

Isentropic efficiency of nozzle can be define as the ratio of actual enthalpy drop put to the Ideal or isentropic enthalpy drop.

[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]

Solution:

Given:

Change in entropy of the system, ΔS = 12J/K

Temperature of the system, [tex]T_{o}[/tex] = 250K

Now, we know that the change in entropy of a system is given by the formula:

ΔS = [tex]\frac{\Delta Q}{T_{o}}[/tex]

Amount of heat added, ΔQ = [tex]\Delta S\times T_{o}[/tex]

ΔQ = 3000J

Answer:

0.1788 ,180°,90°,60°

Explanation:

CONVERSION FROM DEGREE TO RADIANS: For converting degree to radian we have to multiply with [tex]\frac{\pi}{180}[/tex]

using this concept 10.25°=10.25×[tex]\frac{\pi}{180}[/tex]=0.1788

CONVERSION FROM RADIAN TO DEGREE: For converting radian to degree we have to multiply with[tex]\frac{180}{\pi}[/tex]

using this concept π=π×[tex]\frac{180}{\pi}[/tex]

                                  =180°

  [tex]\frac{\pi}{2}[/tex]= [tex]\frac{\pi}{2}[/tex×[tex]\frac{180}{\pi}[/tex]

                                    =90°

  [tex]\frac{\pi}{3}[/tex]= [tex]\frac{\pi}{3}[/tex]×[tex]\frac{180}{\pi}[/tex]

                                    =60°

Answer:

10.25° = 0.1790 radians

π radians = 180°

π/2 radians = 90°

π/3 radians = 60°

Explanation:

The conversion of degree into radians is shown below:

1° = π/180 radians

So,

10.25° = (π/180)*10.25 radians

Also, π = 22/7

So,

[tex]10.25^0=\frac{22\times10.25}{7\times180}radians[/tex]

Solving it we get,

10.25° = 0.1790 radians

The conversion of radians into degree is shown below:

1 radian = 180/π°

(a)

π radians = (180/π)*π°

Thus,

π radians = 180°

(b)

π/2 radians = (180/π)*(π/2)°

[tex]\frac {\pi }{2} radians=\frac{180}{\not {\pi }} \times \frac{\not {\pi }}{2}^0[/tex]

π/2 radians = 90°

(c)

π/3 radians = (180/π)*(π/3)°

[tex]\frac {\pi }{3} radians=\frac{180}{\not {\pi }} \times \frac{\not {\pi }}{3}^0[/tex]

π/3 radians = 60°

Answer: False

Explanation: No, brass is not a ferrous alloy.  

      Ferrous alloys are those alloy which contain iron like cast iron, steel, strain-less steel, high carbon steel. Brass on the other hand does not contain any composition. of iron hence it can not be considered as a ferrous alloy. Brass comes under the category of non- ferrous made with a composition of copper and zinc, however their proportion is not strict and we can add other elements like aluminium or lead to alter its durability or corrosiveness.  

Answer with Explanations:

We are given:

a(t)=4*t^2-2............................(1)

where t= time in seconds, and a(t) = acceleration as a function of time.

and

x(0)=-2 .................................(2)

x(2) = -20 ............................(3)

where x(t) = distance travelled as a function of time.

Need to find x(4).

Solution:

From (1), we express x(t) by integrating, twice.

velocity = v(t) = integral of (1) with respect to t

v(t) = 4t^3/3 - 2t + k1     ................(4)

where k1 is a constant, to be determined.

Integrate (4) to find the displacement x(t) = integral of (4).

x(t) = integral of v(t) with respect to t

= (t^4)/3 - t^2 + (k1)t + k2  .............(5)   where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x(0) = 0 - 0 + 0 + k2 = -2  => k2 = 2   ......................(6)

substitute (6) in (3)

x(2) = (2^4)/3 - (2^2) + k1(2) -2  = -20

16/3 -4 + 2k1 -2 = -20

2k1 = -20-16/3 +4 +2 = -58/3

=>

k1 = -29/3  ....................................(7)

Thus substituting (6) and (7) in (5), we get

x(t) = (t^4)/3 - t^2 - 29t/3 + 2   ..............(8)

which, by putting t=4 in (8)

x(4) = (4^4)/3 - (4^2 - 29*4/3 +2

= 86/3, or

= 28 2/3, or

= 28.67 (to two places of decimal)

The answer is "28.87 m" and the further calculation can be defined as follows:

Given:

[tex]\to a=(4t^2-2)\ \frac{m}{s}\\\\ [/tex]

When

[tex] \to t=0 \ \ \ \ \ \ \ v= 2\ m\\\\ \to t=2 \ \ \ \ \ \ \ v= 20\ m\\\\ \to t=4\ \ \ \ \ \ \ v=?[/tex]

To find:

value=?

Solution:

[tex]V(t)=\int a(t)\ dt=\int (4t^2-2)\ dt=\frac{4}{3}t^3-2t+C_1\\\\ x(t)=\int V(t)\ dt =\int (\frac{4}{3}t^3-2t+C_1)\ dt=\frac{1}{3}t^4-t^2+C_1t+C_2\\\\ x(0)=-2=\frac{1}{3} 0^4-0^2+C_1(0)+C_2\to C_2=-2\ m\\\\ x(2)=-20=\frac{1}{3} 2^4-2^2+C_1(2)-2\to C_1=-\frac{29}{3}\ \frac{m}{s}\\\\ x(4) = \frac{1}{3}4^4-4^2 -\frac{29}{3}4-2 = 28.87 \ m \\\\[/tex]

The particle will be at 28.87 m at the right of the origin.

Learn more about velocity:

brainly.com/question/24376522

Answer:

30

Explanation:

Legally that's when you have to respond

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have [tex]volume of box = 25ft*10ft*12ft= 3000ft^{3}[/tex]

[tex]Volume= 84.95m^{3}[/tex]

[tex]Since 1ft^{3} =0.028m^{3}[/tex]

Now the weight of water displaced = [tex]Weight =\rho \times Volumewhererho[/tex] is density of water = 1000kg/[tex]m^{3}[/tex]

Thus weight of liquid displaced = [tex]\frac{84.95X1000}{1000}tonnes=84.95 tonnes[/tex]..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

Answer: d) All of the above

Explanation: Crystallization is the process in which a solid crystalline structured material is obtained from the liquid substance. the quantity of crystallization center in crystal may increase due to several reasons like changing the cooling rate or mechanical mixing of the substances or imitation of crystals etc.

These processes end up adding mineral atoms to get attached to the center of crystal and hence increasing the size. Thus, the correct option is option (d).

Answer:

Isolated system

Explanation:

By definition of a closed system it means that a system that does not interact with it's surroundings in any manner

The other options are explained as under:

Isothermal system : It is a system that does not allow it's temperature to change

Control Mass system : It is a system whose mass remains conserved which means the mass entering the system equals the mass leaving the system

Open system: It is a system that allows transfer of mass and energy across it's boundary without any opposition i.e freely.

Answer:

Amplitude of A is 4.975 mm and total force is 94.3 N

Explanation:

given data in question

mass (m) = 2 kg

stiffness (k) = 15 kN/m

viscous damping (c) = 5Ns/m

amplitude (F) = 25 N

angular frequency (ω) = 100 rad/s

to find out

amplitude of the forced  and maximum force transmitted

Solution

static force for transmitted is mg i.e 2 × 9.81 = 19.6 N .............. 1

we know the amplitude formula i.e.

Amplitude of A = amplitude /   [tex]\sqrt{c^{2}\omega^{2} + (k - m \omega^{2})^{2}[/tex]

now put the value c k m and ω and we find amplitude

Amplitude of A = 25 /   [tex]\sqrt{5^{2} * 100^{2} + (15000 - 2 * 100^{2})^{2}[/tex]

Amplitude of A = 4.975 mm

now in next part we know the maximum force value when amplitude is equal displacement i.e.

maximum force = amplitude of A [tex]\sqrt{k^{2}+c^{2}\omega^{2}}[/tex]

now put all these value c , ω k and amplitude and we get

maximum force = 4.975 [tex]\sqrt{15000^{2}+5^{2} * 100^{2}}[/tex]

maximum force = 74.7 N                          .......................2

total force is combine equation 1 and 2 we get

total force 19.6 + 74.7 = 94.3 N

Answer:

(a) air

Explanation:

in the following given quenching materials air is the least serve quenching action quenching is a process of heating the material and then rapidly cooling it. Quenching freezes the structure of the material including stresses. Oil has the most sever in its quenching action the commonly used used oil for quenching process is peanut and canola oil