The number-average molecular weight of a poly(styrene-butadiene alternating copolymer is 1,350,000 g/mol; determine the average number of styrene and butadiene repeat units per molecule.

Answer: Covalent bond

Explanation: Covalent bond is the bond that gets created when there is a sharing of electrons among atoms  and hence creating atomic interaction. The bond formed is from the shared pair because they allow the atoms or ions to achieve stability by completely filling the outer shell of the electron and thus form the covalent bond .Therefore, the correct option is the option(b) .

Answer: 5.35m

Explanation:

By using energy equation:

[tex]\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g}  =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}[/tex]

[tex]\gamma=specific weight[/tex]

[tex]v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s[/tex]

[tex]h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}[/tex]

[tex]h_{L}=\frac{(8-7.3)\times 100 }{9.81}  +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}[/tex]

[tex]h_L=7.135+3.927\\h_L=11.062m[/tex]

exit velocity head = [tex]\frac{v_2^{2} }{2g}[/tex]=2.068m

head loss as a function of exit velocity head is=[tex]\frac{11.062}{2.068}[/tex]

[tex]h_L=K\times V_e[/tex]

head loss as a function of exit velocity head =5.35m

Answer: Laplace equation provides a linear solution and helps in obtaining other solutions by being added to various solution of a particular equation as well.

Inviscid , incompressible and irrotational field have and basic solution ans so they can be governed by the Laplace equation to obtain a interesting and non-common solution .The analysis of such solution in a flow of Laplace equation is termed as potential flow.

Answer:

0 K

Explanation:

The thermodynamic absolute temperature is that temperature  at which there is an infinite cooling and so there is no movement of any molecules or particles it is given by kelvin. 0 K is that temperature at which there is no movement of molecule so 0 K or -273°C in degree Celsius is the absolute temperature in thermodynamic temperature scale.  

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

[tex] A=\pi*D^2/4 [/tex]

Replacing the diameter the area results:

[tex] A= 17.76 in^2 [/tex]

Therefore the the stress results:

[tex] σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi [/tex]

Answer:

3/4 th fraction of the volume of block is submerged in oil.

Explanation:

We know that

density of the block, ρ[tex]_{b}[/tex] =SG[tex]\times[/tex]density of water

                                                           = 0.6[tex]\times[/tex] 1000

                                                           = 600 kg/[tex]m^{3}[/tex]

density of the oil, ρ[tex]_{o}[/tex] =SG [tex]\times[/tex] denity of water

                                                           = 0.8[tex]\times[/tex] 1000

                                                           = 800 kg/[tex]m^{3}[/tex]

Let acceleration due to gravity, g = 9.81 m/[tex]s^{2}[/tex]

Volume of block submerged in oil is [tex]V_{1}[/tex]

Volume of block above the oil surface is [tex]V_{2}[/tex]

The total volume of the block is V = [tex]V_{1}[/tex]+[tex]V_{2}[/tex]

Therefore for a partially submerged body, we know that

Buoyant force = total weight of the block

and

total weight of the block = Weight of the fluid displaced by the block

ρ[tex]_{b}[/tex]\timesV\timesg = ρ[tex]_{o}[/tex]\times[tex]V_{1}[/tex]\times g

600([tex]V_{1}[/tex]+[tex]V_{2}[/tex]) = 800 [tex]V_{1}[/tex]

600[tex]V_{1}[/tex] + 600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex]

600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex] - 600[tex]V_{1}[/tex]

600[tex]V_{2}[/tex] = 200[tex]V_{1}[/tex]

[tex]\frac{V_{1}}{V_{2}}[/tex] = [tex]\frac{600}{200}[/tex]

Thus [tex]V_{1}[/tex] = 600

        [tex]V_{2}[/tex] = 200

        V = 600+200

            = 800

Therefore fraction of volume of the block submerged in oil is,

       [tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{600}{800}[/tex]

       [tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{3}{4}[/tex]

Answer:

Explained

Explanation:

Cold working: It is plastic deformation of material at temperature below   recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.

Given melting point temp of lead is 327° C and lead recrystallizes at about

0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.

Answer:

t =253.8s

Explanation:

Chvorinov's Rule can be written as:

[tex]t=B(\frac{V}{A} )^{n}[/tex]

where t is the solidification time,

V is the volume of the casting,

A is the surface area of the casting that contacts the mold,

n is a constant

B is the mold constant

The S.I. units of the mold constant B are s/m2.

According to Askeland, the constant n is usually 2.

[tex]V=\frac{\pi D^{2}h }{4} = 3.9*10^6[/tex]mm3

[tex]As=\pi D h+2\frac{\pi }{4} D^{2} =0.424*10^6[/tex] mm2

[tex]V/A=9.198[/tex]mm

[tex]t = 3.0*9.198^2[/tex] =253.8s

Answer:

Chvorinov's Rule with Askeland Method: t = 4.286694102 minutes

Chvorinov's Rule with Degarmo Method:

Explanation:

Data:

Aluminum disc

Diameter (D) = 500 mm

Thickness = Height (h) = 20 mm

Mold Constant (C) = 3.0 sec / [tex]mm^{2}[/tex]

Required:

Solidification time (t) in minutes = ?

Formula:

The solidification time can be found by using the Chvorinov's Rule:

[tex]t = C (\frac {V}{A})^{n}[/tex]

Where;

t = solidification time

C = mold constant

V = Volume of disc

A = Surface area of disc

n = constant

Note: According to Askeland n = 2.0 and According to Degarmo n varies 1.5 to 2.0 therefore , we will do for both method and by Degarmo method we can predict maximum and minimum solidification time.

Solution:

First, we will find the volume of the disc

disc = cylinder

therefore, Volume of cylinder is given by:

[tex]V = \frac{\pi }{4} * D^{2} * H[/tex]

Where:

V = Volume of Cylinder

H = Height of disc

D = Diameter of disc

Now, putting dimensional values in above equation

[tex]V = \frac{\pi }{4} * 500^{2}  *20[/tex]

V = 3926990.817 [tex]mm^{3}[/tex]

Second, we will find the surface area of the disc

Therefore, surface area of cylinder is given by:

[tex]A = (\pi * D * H) + (2 * \frac{\pi }{4} * D^{2} )[/tex]

Where:

A = Surface area of disc

D = Diameter of disc

H = Height of disc

Now, putting dimensional values in above equation

[tex]A = (\pi * 500 * 20) + (2 * \frac{\pi }{4} * 500^{2} )[/tex]

A = 424115.0082 [tex]mm^{2}[/tex]

Finally, Moving towards the final solution

Rewriting the equation:

[tex]t = C (\frac {V}{A})^{2}[/tex]

Putting the dimensional and constants values in the equation

[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]

t = 257.2016461 seconds

Converting to minutes

t = 4.286694102 minutes

Rewriting the equation:

[tex]t = C (\frac {V}{A})^{2}[/tex]

Putting the dimensional and constants values in the equation

[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{1.5}[/tex]

t = 84.52508604 seconds

Converting to minutes

t = 1.408751434 minutes

Rewriting the equation:

[tex]t = C (\frac {V}{A})^{2}[/tex]

Putting the dimensional and constants values in the equation

[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]

t = 257.2016461 seconds

Converting to minutes

t = 4.286694102 minutes

Answer:  b) FCOR( Face-centered orthorhombic)

Explanation: Face centered orthorhombic lattice is the lattice that has eight lattice corner points and they also have each face  with center lattice point.This lattice structure is usually not common in metals.Whereas the face centered lattice structure , body structure lattice and hexagonal close packed lattice are common in metal.Thus option (b) is the correct option.

Solution:

Given :

atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m

atomic wieght, M = 95.96

Cell structure is BCC (Body Centred Cubic)

For BCC, we know that no. of atoms per unit cell, z = 2

and atomic radius, r =[tex]\frac{a\sqrt{3} }{4}[/tex]

so, a = [tex]\frac{4r}{\sqrt{3}}[/tex]

m = mass of each atom in a unit cell

mass of an atom = [tex]\frac{M}{N_{A} }[/tex],

where, [tex]N_{A}[/tex] is Avagadro Number = 6.02×10^{23}

volume of unit cell = a^{3}

density, ρ = [tex]\frac{mass of unit cell}{volume of unit cell}[/tex]

density, ρ = [tex]\frac{z\times M}{a^{3}\times N_{A}}[/tex]

ρ = [tex]\frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}[/tex]

ρ = 10.215gm/[tex]cm^{3}[/tex]

Answer:

a). TRUE

Explanation:

Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.

Mathematically, thermal efficiency is

        η =  net work done / heat input

While heat rate is  the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.

Mathematically, heat rate is

       Heat rate = heat input / net work done

Thus from above we can see that heat rate is the reciprocal of thermal efficiency.

Thus, Heat rate is reciprocal of thermal efficiency.

Answer:

ΔK.E. = - 142.72 kJ

Explanation:

mass = 884 kg

initial velocity = 68 km/h = 68 \times \frac {5}{18} = 18.89 m/s

final velocity = 0 m/s

height = 69 m

change in kinetic energy :

ΔK.E. = [tex]\dfrac{1}{2}m(v_f^2-v_i^2)[/tex]

ΔK.E. =[tex]\dfrac{1}{2}\times 884 \times (0^2-18.89^2)[/tex]

ΔK.E. =-142,716.05 J

ΔK.E. =-142.72 kJ

hence change in  kinetic energy of the automobile is  -142.72 kJ

Answer:

The Answer to the question is :

Explanation:

The contact angle between the mercury surface and capillary tube wall is Greater than 90.

If the surface of the solid is hydrophobic, the contact angle will be greater than 90 °. On very hydrophobic surfaces the angle can be greater than 150º and even close to 180º.

Answer:

a) [tex]h_L=1.17m[/tex]

b)[tex]h_L=1.52m[/tex]

c)[tex]P_1=1.721 kN[/tex]

  [tex]P_2=2.236 kN[/tex]

Explanation:

velocities of the pipe;

velocity of small dia pipe

[tex]v_{small}=\frac{Q}{A_{small}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.15^2 }=8.52m/s[/tex]

velocity of larger dia pipe

[tex]v_{large}=\frac{Q}{A_{large}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.25^2 }=3.05m/s[/tex]

a) pressure head loss when water is flowing from large to smaller pipe

  [tex]h_L=(\frac{1}{C_c} -1)^2 \times \frac{v^2}{2g}[/tex]

[tex]h_L=(\frac{1}{0.64} -1)^2 \times \frac{8.52^2}{2\times 9.81} = 1.17m[/tex]

b) pressure head loss when water flow from small pipe to large pipe

[tex]h_L= \frac{v^2}{2g}(1-\frac{A_{small}}{A_{large}} )^2[/tex]

[tex]h_L=\frac{8.52^2}{2\times9.81}(1-\frac{\frac{\pi}{4}\times 0.15^2}{\frac{\pi}{4}\times 0.25^2}} )^2= 1.52 m[/tex]

c) power loss in both cases are

[tex]P_1= \rho g Q h_{L1}= 1000 \times 9.81\times 0.15\times 1.17= 1.721 kN[/tex]

[tex]P_2= \rho g Q h_{L2}= 1000 \times 9.81\times 0.15\times 1.52= 2.236 kN[/tex]

Explanation:

Closed die forging:

This is also known as impression die forging.By using the two die ,impression are made to produce desired forge product.A small gap provides between these two dies ,this small gap is called flash gutter.By the help of flash gutter excess ,material can flow and forms flash.

This flash plays an important role in the closed die forging.Generally friction is high in the flash gutter due to high ratio of length to thickness.So material feels high pressure and confirm the filling of die cavity.

Answer:

output=[tex]\frac{2}{(s+3)(s+4)}[/tex]

Explanation:

output =transfer function ×input

here transfer function G=[tex]\frac{2}{(S+3)(S+4)} {}[/tex]

input = unit impulse

in S domain unit impulse =1

so output =[tex]\frac{2}{(S+3)(S+4)} {}[/tex]

=[tex]\frac{2}{(s+3)(s+4)}[/tex]

Answer:

Design of Experiment also known as DoE is a systematic methodology to understand how any process or parameters of any product affects the response variables like physical properties or performance of any product, etc. It serves the purpose of making the job easier.

It is technique to generate valuable information required with minimum experimentation with the use of these:

Process Optimization:

It includes the following:

Answer:

D Is the answer babe....

Answer:

d)- ON/OFF.

Explanation:

ON/OFF components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated.

Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   [tex]Pr=\dfrac{\nu }{\alpha }[/tex]

  Where [tex]\nu[/tex] is the molecular diffusivity of momentum

             [tex]\alpha[/tex] is the molecular diffusivity of heat.

 Prandtl number(Pr) can also be defined as

    [tex]Pr=\left (\dfrac{\delta }{\delta _t}\right )^3[/tex]

Where [tex]\delta[/tex] is the hydrodynamic boundary layer thickness and [tex]\delta_t[/tex] is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

Answer:

true

Explanation:

Creep is known as the time dependent deformation of structure due to constant load acting on the body.

Creep is generally seen at high temperature.

Due to creep the length of the structure increases which is not fit for serviceability purpose.

When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.

When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.

The answer is c) atomic number

Answer: 5) KC-0.78 KU

Explanation:

 KC-0.78 KU is not defined for tuning of overshoot on the Start-up as, this method is only applicable for there is some turning constants and also there is no overshoot during the normally modulating control. But some overshoot at start up are applicable. For the continuous cycling method, some overshoot is same as for closed loop tuning.    

Answer: False

Explanation:

Pure silicon is a good semiconductor, which conducts electricity when is mixed with some other component and can also work as an insulator.It is found abundantly on the earth's crust area. It is widely used in mixed form in the advanced electronic technologies but not in pure silicon form . So silicon is the important material in the advanced electronic technologies but not the pure silicon form.

Answer:

Screw pumps like those that are used in viscous fluids transportation have lubricating properties. The fluid flows in a line axially without any disturbence. Since, the motion of fluid is free from any sort of rotation even at high speeds there is no turbulence and motion of the pump is quite. Thus screw Pumps provides smooth operation with extremely low pulsation, lower noise levels and higher efficiency.

Answer:  

  Degree of crystallinity affects polymer properties as increasing in   crystallinity means higher the thermal stability and harder the material. Basically, crystallinity strongly affects polymer properties as it defines the degree of long range order in a material. Crystallinity is also determined by the size as well as molecular chain orientation.

Explanation:

a). Operational Performance

     It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.

b).  Salvage externalities

   Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.

c). Value Vs Risk

   When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.

d). Initial expenditure

   Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.

e). Maintenance implications

   It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.

Answer:

 Is required a 0.8 inches diameter steel shaft.

Explanation:

With the power P and the rotating speed n (RPM), we can find the torque applied:

T = P/N

Before calculating the torque, we convert the power and rotating speed units:

[tex]P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s} [/tex]

[tex]n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}[/tex]

Replacing the values, the torque obtained is:

[tex]T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in[/tex]

Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:

[tex]Smax =\ \frac{T.R}{J}[/tex]

Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:

[tex]J =\frac{\pi.D^4}{32}[/tex]

Where D is shaft's diameter. Replacing the expression of J in

[tex]Smax =\frac{T.R}{\frac{\pi.D^4}{32}}[/tex]

As the radius is half of the diameter:

[tex]Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}[/tex]

For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:

[tex]Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}[/tex]

Solving for D:

[tex]D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in[/tex]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

Answer:

0.424

Explanation:

The electrical energy is the electrical power times time:

E = Pt

E = (IV)t

E = (1.5 A) (110 V) (300 s)

E = 49,500 J

The heat absorbed by the cookie dough is the mass times specific heat capacity times increase in temperature:

Q = mCΔT

Q = (1 kg) (4200 J/kg/K) (5 K)

Q = 21,000 J

So the fraction of electrical energy converted to internal energy is:

Q / E = 21,000 / 49,500

Q / E = 0.424

The remainder of the electrical energy is used to do work.

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

[tex]X_L=0.2\times 12=2.4[/tex] ohm

[tex]X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm[/tex]

[tex]Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}[/tex]

[tex]Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}[/tex]

=5.021 ohm

so amplitude of current =  [tex]\frac{v}{z}=\frac{120}{5.021}=23.89[/tex]