What is the definition of diameter pitch?

Answer:

b)false

Explanation:

Specific cutting energy:

 Energy required to remove unit volume of material is called specific energy.In other words we can say that the ratio of energy to the volume removal rate is called specific cutting energy.

When cutting speed is increases then specific energy goes to decrease.As well as when depth of cut and feed of tool is increase then specific cutting energy will decrease.

Answer:

The given statement is true.

Explanation:

A chemical property of any material be it solid liquid or gas is defined as how it interacts chemically with other substances after an interaction takes place between them.  The interaction in language of chemistry is known as chemical reaction. Different materials in nature show different interactions with various other substances.The 2 substances can react with each other and form a different compound or may not react with each other and are termed as inert chemicals. Infact it is this interaction between the various chemicals that we can group different into classes based on their behavior with different chemicals. The interaction of different materials act as their signatures that help us in identifying them.

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = [tex]\frac{60+180}{2}[/tex]  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

Answer:

the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]

Explanation:

Given that

Thermal conductivity K=14.54 W/m.K

This above given conductivity is in SI unit.

     SI unit                                           IP unit              Conversion factor

    m                                                      ft                      0.3048

   W                                                       Btu/hr               0.293          

The unit of conductivity in IP is Btu./ft.hr.F.

Now convert into IP divided by 1.73 factor.

[tex]0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}[/tex]

So

[tex]0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]

[tex]8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]

So the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]

Answer:

Rated power = 1345.66 W/m²

Mechanical power developed = 3169035.1875 W

Explanation:

Wind speed, V = 13 m/s

Coefficient of performance of turbine, [tex]C_p[/tex] = 0.3

Rotor diameter, d = 100 m

or

Radius = 50 m

Air density, ρ = 1.225 kg/m³

Now,

Rated power = [tex]\frac{1}{2}\rho V^3[/tex]

or

Rated power = [tex]\frac{1}{2}\times1.225\times13^3[/tex]

or

Rated power = 1345.66 W/m²

b) Mechanical power developed =  [tex]\frac{1}{2}\rho AV^3C_p[/tex]

Here, A is the area of the rotor

or

A = π × 50²

thus,

Mechanical power developed = [tex]\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3[/tex]

or

Mechanical power developed =  3169035.1875 W

Answer:

10.41 kg

Explanation:

The gas state equation is:

p * V = n * R * T

For this equation we need every value to be in consistent units

1.5 bar = 150 kPa

5 C = 278 K

n = p * V / (R * T)

n = 150000 * 10 / (8.31 * 278) = 649 mol

Multiplying the amount of moles by the molecular weight of the gas we obtain the mass:

m = M * mol

m = 16.04 * 649 = 10410 g = 10.41 kg

Answer:

Mass of oil will be 7.176 pound

Explanation:

We have given specific gravity of oil  = 0.86

We know that specific gravity is given by [tex]specific\ gravity=\frac{density\ of\ oil}{density\ of\ water}[/tex]

[tex]0.86=\frac{density\ of\ oil}{1000}[/tex]

Density of oil = [tex]860kg/m^3[/tex]

We have given volume of oil = 1 gallon

We know that 1 gallon = 0.003785 [tex]m^3[/tex]

So mass of oil = volume ×density

mass = 0.003785×860 = 3.2551 kg

We know that 1 kg = 2.2046 pound

So 3.2551 kg = 3.2551×2.2046 = 7.176 pound

Answer:

D/t>20

Explanation:

Lets take

D =Diameter of thin cylinder

t =Thickness of thin cylinder

So a cylinder is called thin cylinder if the ratio of diameter to the thickness is greater than 20 (D/t>20 ).

But on the other hand a cylinder is called thick cylinder is ratio of  thickness to the diameter is greater than 20 (t/D>20 ).

So the governing ratio of thin walled cylinder is 20.

Answer:

[tex]specific\ volume=0.00097\ m^3/kg[/tex]

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 [tex]kg/m^3[/tex]

Density of sea water = 1025  [tex]kg/m^3[/tex]

We know that

[tex]Density=\dfrac{mass}{Volume}[/tex]   ---1

Specific volume

[tex]specific\ volume=\dfrac{Volume}{mass}[/tex]    ---2

From equation 1 and 2

We can say that

[tex]specific\ volume=\dfrac{1}{density}\ m^3/kg[/tex]

[tex]specific\ volume=\dfrac{1}{1025}\ m^3/kg[/tex]

[tex]specific\ volume=0.00097\ m^3/kg[/tex]

Answer:

The change of entropy is 1.229 kJ/(kg K)

Explanation:

Data

[tex] T_1 = 300 K [/tex]

[tex] T_2 = 900 K[/tex]

[tex] p_1= 400 kPa[/tex]

[tex] p_2= 300 kPa[/tex]

[tex] R= 0.287 kJ/(kg K)[/tex] (Individual Gas Constant for air)  

For variable specific heats  

[tex]s(T_2, p_2) - s(T_1, p_1) = s^0(T_2) - s^0(T_1) - R \, ln \frac{p_2}{p_1}[/tex]

where [tex] s^0(T) [/tex] is evaluated from table  attached

[tex] s^0(900 K) = 2.84856 kJ/(kg K)[/tex]

[tex] s^0(300 K) = 1.70203 kJ/(kg K)[/tex]

Replacing in equation

[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 2.84856 kJ/(kg K) - 1.70203 kJ/(kg K) - 0.287 kJ/(kg K) \, ln \frac{300 kPa}{400 kPa}[/tex]

[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 1.229 kJ/(kg K)[/tex]

Explanation:

The two types of furnaces used in steel production are:

Basic oxygen furnace

In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.

Electric arc furnace

Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.

Answer:

1) The change in storage of the catchment is 707676800 cubic meters.

2) The runoff coefficient of the catchment is 0.83.

Explanation:

The water budget equation of the catchment can be written as

[tex]P+Q_{in}=ET+\Delta Storage+Q_{out}+I[/tex]

where

'P' is volume of  precipitation in the catchment =[tex]Area\times Precipitation[/tex]

[tex]Q_{in}[/tex] Is the water inflow

ET is loss of water due to evapo-transpiration

[tex]\Delta Storage[/tex] is the change in storage of the catchment

[tex]Q_{out}[/tex] is the outflow from the catchment

I is losses due to infiltration

Applying the values in the above equation and using the values on yearly basis (Time scale is taken as 1 year) we get

[tex]4000\times 10^{6}\times 1.02+0=0.40\times 4000\times 10^{6}+\Delta Storage+34.2\times 3600\times 24\times 365\times 5.5\times 10^{-9}\times 4000\times 10^{6}\times 3600\times 24\times 365[/tex]

[tex]\therefore \Delta Storage=707676800m^3[/tex]

Part b)

The runoff coefficient  C is determined as

[tex]C=\frac{P-I}{P}[/tex]

where symbols have the usual meaning as explained earlier

[tex]\therefore C=\frac{102-5.5\times 10^{-7}\times 3600\times 24\times 365}{102}=0.83[/tex]

Answer:

Shape factor

Explanation:

Shape factor is the ratio of maximum plastic moment to maximum elastic moment.Shape factor is denoted by K.

Shape factor can be given as

[tex]K=\dfrac{M_p}{M_y}[/tex]

[tex]K=\dfrac{\sigma _yZ_p}{\sigma _y Z}[/tex]

[tex]K=\dfrac{Z_p}{ Z}[/tex]

For a solid rectangular section made from ductile material shape factor is 1.5 .

Answer:

The mass of the air is 4.645 kg.

The work done or heat transfer is 277.25 kj.

Explanation:

In isothermal process PV=constant. Take air as an ideal gas. For air gas constant is 287 j/kgK.

Given:  

Initial pressure of the gas is 2 bar.

Initial volume of the gas is 2 m³.

Initial temperature is 300 K.

Final pressure is 1 bar.

Calculation:  

Step1

Apply ideal gas equation for air as follows:

PV=mRT

[tex]2\times10^{5}\times2=m\times 287\times300[/tex]

m = 4.645 kg.

Thus, the mass of the air is 4.645 kg.

Step2  

For isothermal process work done is same as heat transfer.

Work done or the heat transfer is calculated as follows:

[tex]W=P_{i}V_{i}ln\frac{P_{i}}{P_{f}}[/tex]

[tex]W=2\times10^{5}\times2\times ln\frac{2}{1}[/tex]

[tex]W=2.7725\times10^{5}[/tex] j

Or,

W=277.25 kj.

Thus, the work done or heat transfer is 277.25 kj.  

Answer:

1.2253 kg/m3

Explanation:

from ideal gas equation we  know that

[tex]PV = n\timesRT[/tex]

[tex]\frac{n}{V} = \frac{P}{(RT)}[/tex]

where,

pressure P = 1.01325 x 10^5 N/m2 = 101325 Pascals  

gas constant R = 8.314 J/mol/K  

T = 288.15 K

[tex]\frac{n}{V} = \frac{101325}{(8.314*288.15K)}[/tex]

[tex]\frac{n}{V} = 42.2948967 Pa*mol/J[/tex]

we know that 1 Pa = 1 J/m3

so[tex]\frac{n}{V} = 42.2948967\  mol/m3[/tex]

as we know, Air consist of  21% oxygen and 79% nitrogen

therefore

Molar mass of air:

[tex]0.21 *(32 g/mol) + 0.79 *(28 g/mol) = 28.97 g/mol[/tex]

So  [tex](42.2948967\  mol/m3) \times 28.97\  g/mol[/tex]

= 1 225.27 g/m^3

= 1.22528316 kg/m^3 ~ 1.2253 kg/m3

Answer:

The tool's trajectory in a CAM program refers to the places where the tool will be during the work. It is important to review it before generating the program for the following reasons

1. analyze the machining strategy and identify which one is better for each piece.

2.Avoid the collision of the tool holder with the work piece.

3.Avoid the shock of the tool with the piece.

4. Prevent the collision of the tool with elements that are not displayed on the CAM such as clamping flanges or screws.

Answer:

74.2 km from station A.

Explanation:

We set a frame of reference with the station at the origin and the X axis pointing in the direction the trains run.

The freight train leaves at 8 AM and travels 10 minutes at 45 km/h.

For this problem it is better to convert the speeds to km/min

45 km/h = 0.75 km/min

The equation for position under constant speed is:

X(t - t0) = X0 + v0 * (t - t0)

Since we know the time it will stop moving at this speed:

X(10 - 0) = 0 + 0.75 * (10 - 0) = 7.5 km

After it ran those 7.5 km it will keep running at 60 km/h.

60 km/h = 1 km/min

The position equation for it is now:

X(t - 10) = 7.5 + 1 * (t - 10)

The passenger train leaves the station at 8:35 AM. It travels at 75 km/h for 5 minutes.

75 km/h = 1.25 km/min

After those 5 minutes it will have traveled:

X(40 - 35) = 0 + 1.25 * (40 - 35) = 6.25 km

Then it travels at 105 km/h

105 km/h = 1.75 km/min

Its position equation is now:

X(t - 40) = 6.25 + 1.75 * (t - 40)

Equating both positions we find the time at which they would meet:

7.5 + 1 * (t - 10) = 6.25 + 1.75 * (t - 40)

7.5 + t - 10 = 6.25 + 1.75*t - 70

t - 1.75*t = 6.25 - 70 +10 - 7.5

-0.75*t = -61.25

t = 61.25 / 0.75

t = 81.7 minutes

The freight train will have to be parked 5 minutes before this at t = 76.7 minutes.

At that moment the freight train will be at:

X(76.7 - 10) = 7.5 + 1 * (76.7 - 10) = 74.2 km

Answer:

Time taken to reach 80 meters equals 4.4897 seconds.

Explanation:

We know that velocity is related to position as

[tex]v=\frac{ds}{dt}[/tex]

Now it is given that [tex]v=(11+0.2v[/tex]

Using the given velocity function in the above relation we get

[tex]\frac{ds}{dt}=(11+0.2s)\\\\\frac{ds}{(11+0.2s)}=dt\\\\\int \frac{ds}{(11+0.2s)}=\int dt\\\\[/tex]

Now since the limits are given as

1) at t = 0 , s=0

Using the given limits we get

[tex]\int_{0}^{80} \frac{ds}{(11+0.2s)}=\int_{o}^{t} dt\\\\\frac{1}{0.2}[ln(11+0.2s)]_{0}^{80}=(t-0)\\\\5\times (ln(11+0.2\times 80)-ln(11))=t\\\\\therefore t=4.4897seconds[/tex]

Answer:

0.014 in

Explanation:

The Young's module of steel is of E = 210 GPa = 30*10^6 psi

The section of the bolt would be:

A = π/4 * D^2

The stiffness would be:

k = E * A / L

k = π * D^2 * E / (4 * L)

k = π * 0.25^2 * 30*10^6 / (4 * 10) = 147000 lb / in

If I apply a force of 2000 lb, I calculate the  stretching with Hooke's law:

Δx = f / k

Δx = 2000 / 147000 = 0.014 in

Answer:

force R = 846.11 N

lifting force L = 110.36 N

if cable fail complete both R and L will be zero

Explanation:

given data

mass woman mw = 60 kg

mass package mp = 9 kg

accelerates rate a = g/4

to find out

force R and lifting force L and if cable fail than what values would R and L acquire

solution

we calculate here first reaction R force

we know elevator which accelerates upward

so now by direction of motion , balance the force that is express as

R - ( mw + mp ) × g = ( mw + mp ) × a

here put all these value and a = g/4 and use g = 9.81 m/s²

R - ( 60 + 9 ) × 9.81 = ( 60 + 9  ) × g/4

R = ( 69  ) × 9.81/4  + ( 69 ) 9.81

R = 69  ( 9.81 + 2.4525 )

force R = 846.11 N

and

lifting force is express as here

lifting force = mp ( g + a)

put here value

lifting force = 9 ( 9.81 + 9.81/4)

lifting force L = 110.36 N

and

we know if cable completely fail than body move free fall and experience no force

so both R and L will be zero

Answer:

Boiling point of HF is higher as compared to HCl because of presence of hydrogen bonding in it.

Explanation:

In HF, intermolecular force of attraction is hydrogen bonding.

Hydrogen bonding is a type of electrostatic force of attraction existing between H atom and electronegative atom.

For a molecule to have hydrogen bonding, H atom must be bonded to electronegative atom, O, N and F.

Hydrogen bonding can be intermolecular and intramolecular.

So, in HF hydrogen bonding present.

In HCl, only van der Waals force exists. van der Waals forces are weak as compared to hydrogen bonding.

Because of presence of hydrogen bonding, HF molecules are held tightly and so requires more heat to boil.

Therefore, boiling point of HF is more as compared to HCl.

Answer:

Yes

Explanation:

As we know that heat transfer take place from high temperature body to low temperature body.

In the given problem ,the temperature of the air is high as compare to the temperature  of can of bear ,so the heat transfer will take place from air to can of bear and at the last stage when temperature of can of bear will become to the temperature of air then heat transfer will be stop.Because temperature of the  both body will become at the same  and this stage is called thermal equilibrium.

So an office worker claim is correct.

Answer:

(a) 10.76 [tex]ft^2/sec[/tex]

(b) 0.134 lbm

(c) 0.4308 Btu/lbm

Explanation:

We have to do conversion

(a) conversion of viscosity [tex]1m^2/sec\ to\  ft^2/sec[/tex]

We know that [tex]1m^2=10.76feet^2[/tex]

So [tex]1m^2/sec=10.76ft^2/sec[/tex]

(b) We have to convert power 100 W in hp

We know that 1 W = 0.00134 hp

So 100 W = 100×0.00134=0.134 hp

(c) We have to convert 1 KJ/kg to Btu/lbm

We know that 1 KJ = 0.94780 Btu

And 1 kg = 2.20 lbm

So [tex]1KJ/kg=\frac{0.9478Btu}{2.20lbm}=0.4308Btu/lbm[/tex]

Answer:

Decrease

Explanation:

Generally with increasing the temperature the mechanical properties of material decreases.But some materials have exceptions like tempered steel because when temperature increase then young modulus of elasticity of tempered steel increases.

So we can say that when  with increasing temperature the properties of materials Yield and tensile strengths and modulus of elasticity decreases.

Decrease

Answer:

Decreases

Explanation:

Yield and tensile strength and modulus of elasticity decreases with increasing temperature. As the temperature is increased most materials decrease in their elasticity.

The modulus of elasticity is proportional to its tensile strength.

Answer:

The volume of the extra water is [tex]2.195 ft^{3}[/tex]

Solution:

As per the question:

Mass of the canoe, [tex]m_{c} = 175 lb + w[/tex]

Height of the canoe, h = 21.5 ft

Mass of the kevlar canoe, [tex]m_{Kc} = 38 lb + w[/tex]

Now, we know that, bouyant force equals the weight of the fluid displaced:

Now,

[tex]V\rho g = mg[/tex]

[tex]V = \frac{m}{\rho}[/tex]                                  (1)

where

V = volume

[tex]\rho = 62.41 lb/ft^{3}[/tex] = density

m = mass

Now, for the canoe,

Using eqn (1):

[tex]V_{c} = \frac{m_{c} + w}{\rho}[/tex]

[tex]V_{c} = \frac{175 + w}{62.41}[/tex]

Similarly, for Kevlar canoe:

[tex]V_{Kc} = \frac{38 + w}{62.41}[/tex]

Now, for the excess volume:

V = [tex]V_{c} - V_{Kc}[/tex]

V = [tex]\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}[/tex]

Answer:

Explanation:

Poisson's ration=  is the ratio between the deformation that occurs in the material in the direction perpendicular to the force applied with the deformation suffered by the material in the same direction of the force.

young's modulus= is the ratio between the stress applied to a material with respect to its unit deformation when this material complies with the hooke's law.

shear modulus=change in the way an elastic material experiences when subjected to shear stresses

Answer with Explanation:

Drag and lift are the forces that act on an object which moves in a fluid and are explained as under:

1)Drag: Drag is the force that opposes the motion of an object moving in a fluid and hence is analogous to fluid frictional force in a fluid. Power is needed to be spent by an object to overcome the frictional drag. The drag force is different from the classical friction as we know that the frictional force on a dry surface is independent of the velocity of the object but for drag the force is proportional to the square of the velocity of the object.

Mathematically

[tex]F_{Drag}=\frac{1}{2}C_{d}A\rho _{fluid}V^2[/tex]

where

[tex]C_d[/tex] is a constant known as coefficient of drag and A is the projected area of the object.

2) Lift: Lift force is a component of the force that a fluid exerts on an object moving in it that counters the weight of the object and hence has the tendency to lift the object and hence is known as lift force. The lift force is perpendicular to the incoming fluid. Lift force is useful as it allows the aeroplanes  to fly as the lift force that is generated by the air on the wings of the plane is used to overcome the force of gravity on the plane.

Mathematically

[tex]F_{Lift}=\frac{1}{2}C_{L}A\rho _{fluid}V^2[/tex]

where

[tex]C_L[/tex] is a constant known as coefficient of lift and A is the projected area of the object moving with velocity 'v'.

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is [tex]1.557\times10^{-7}[/tex]m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

[tex]Re=\frac{vd}{\nu}[/tex]

Here, v is velocity, [tex]\nu[/tex] is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

[tex]Re=\frac{vd}{\nu}[/tex]

[tex]Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}[/tex]

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

Answer:

ρ=962.16kg/m^3

Explanation:

The first thing we must do to solve is to find the mass of the specimen using the weight equation

w = mg

m=w/g

m=0.45/9.81=0.04587kg

To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).

We must bear in mind that the principle of archimedes indicates that the buoyant force is given by

F = ρgV

where V is the specimen volume and  ρ is the density of alcohol = 789kg / m ^ 3

considering the above we have the following equation

0.081=0.45-(789)(9.81m/s^2)V

solving for V

V=(0.081-0.45)/(-789x9.81)

V=4.7673x10^-5m^3

finally we found the density

ρ=m/v

ρ=0.04587kg/4.7673x10^-5m^3

ρ=962.16kg/m^3

Answer:

Final temperature will be 16.57°C

Explanation:

We have given mass of nitrogen m = 6 kg

Initial temperature [tex]T_1=30^{\circ}[/tex]

Decrease in internal energy [tex]\Delta U=-60KJ[/tex]

Specific heat of nitrogen [tex]c_v[/tex]= 0.745 KJ/kg

Let final temperature is [tex]T_2[/tex]

Change in internal energy is given by [tex]\Delta U=mc_v\Delta T=mc_v(T_2-T_1)[/tex]

[tex]-60=6\times 0.745(T_2-30)[/tex]

[tex]T_2=16.57^{\circ}C[/tex]

So final temperature will be 16.57°C