Express the dimensions of (a) force, (b) pressure, (c) energy, and (d) power in terms of primary dimensions.

Answer:

this is a typical stress-strain curve for a ductile material

Explanation:

A: proportional limit

B:Elastic limit

C:upper yield point

D:lower yield point

E:ultimate strength

F:rupture strength

Answer:

[tex]V = 0.208 m^3[/tex]

saturated pressure  = 0.91 bar 0r 91 kPa

Explanation:

Given data:

Mass of liquid water 200 kg

Steel tank temperature is 97 degree celcius = 97°C

At T = 97 degree celcius,  saturated pressure  = 0.91 bar 0r 91 kPa

And also for saturated liquid

Specific volume [tex]\nu[/tex] is 0.00104 m^3/kg

Volume of tank is given as

[tex]V = \nu * m[/tex]

[tex]V = 0.00104 * 200[/tex]

[tex]V = 0.208 m^3[/tex]

Measurement error is the difference between observed and calculated values, including both random and systematic errors. Adjusting parameters such as K can help fit experimental data to models, with potential errors addressed through improved methods and investigation of discrepancies.

The difference between the observed and calculated values is known as measurement error or observational error. This error can be classified into two types: random error, which occurs naturally and varies in an unpredictable manner, and systematic error, which is consistent and typically results from a flaw or limitation in the equipment or the experimental design. To identify the sources of these errors, one can compare the experimental data with calculated models and adjust parameters, such as the constant K, to find the best fit.

If the experimental values do not match the accepted values, sources of error should be investigated and procedures modified to improve accuracy. For example, in a physics experiment, one could compare the experimental acceleration to the standard acceleration due to gravity (9.8 m/s²) and identify factors contributing to any discrepancy. Common sources of error might include environmental factors, instrument calibration issues, or procedural mistakes.

The expressions squared or cubed with appropriate metric prefixes are: (a) 184.9 Mg^2, (b) 4 x 10^-18 kg^2, and (c) 12.167 km^3. Metric prefixes are used for clarity in representing large quantities.

The question deals with the operation of squaring and cubing given quantities and expressing them with the appropriate metric prefixes. Let's evaluate each expression step by step:

It is important to use the proper metric prefixes when
expressing large numbers to ensure clarity and avoid confusion.

Answer:

The flow of a real fluid has more complexity as compared to an ideal fluid owing to the phenomena caused by existence of viscosity

Explanation:

For a ideal fluid we know that there is no viscosity of the fluid hence the boundary condition need's not to be satisfied and the flow occur's without any head loss due to viscous nature of the fluid. The friction of the pipe has no effect on the flow of an ideal fluid. But for a real fluid the viscosity of the fluid has a non zero value, the viscosity causes boundary layer effects, causes head loss and also frictional losses due to pipe friction hugely make the analysis of the flow complex. The losses in the energy of the flow becomes complex to calculate as frictional losses depend on the roughness of the pipe and Reynolds number of the flow thus increasing the complexity of the analysis of flow.

Answer:

The fraction of isotope 1 is 50.7 % and fraction  fraction of isotope 1 is 49.2 %.

Explanation:

Given that

Weight of isotope 1 = 78.918 amu

Weight of isotope 2 = 80.916 amu

Average atomic weight= 79.903 amu

Lets take fraction of isotope 1 is x then fraction of isotope 2 will be 1-x.

The total weight will be summation of these two isotopes

79.903 = 78.918 x + 80.916(1-x)

By solving above equation

80.916 - 79.903 = (80.916-78.918) x

   x=0.507

So the fraction of isotope 1 is 50.7 % and fraction  fraction of isotope 1 is 49.2 %.

Answer:

Rt=70.00 ohm

Explanation:

ohm's theory tells us that the connected resistors in series add up directly

Rt=R1+R2+R3+R4.......

Therefore, for this case, the only thing we should do is add the resistance directly

Rt=R1+R2

Rt=40.00 ohm+30.00 Ohm.

Rt=70.00 ohm

the total resistance of this configuration is 70.00 ohm

Answer:

Cut off ratio=2.38

Explanation:

Given that

[tex]T_1=300K[/tex]

[tex]P_1=100KPa[/tex]

[tex]P_2=P_3=7200KPa[/tex]

[tex]T_3=2250K[/tex]

Lets take [tex]T_1[/tex] is the temperature at the end of compression process

For air γ=1.4

[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}[/tex]

[tex]\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}[/tex]

[tex]T_2=1070K[/tex]

At constant pressure

[tex]\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}[/tex]

[tex]\dfrac{2550}{1070}=\dfrac{V_3}{V_2}[/tex]

[tex]\dfrac{V_3}{V_2}=2.83[/tex]

So cut off ratio

[tex]cut\ off\ ratio =\dfrac{V_3}{V_2}[/tex]

Cut off ratio=2.38

Answer:

Thermal resistance:

  Thermal resistance is the property which oppose the the heat transfer.It is the property for measurement of heat flow.

As we know that heat transfer take place from the high temperature to low temperature.Heat transfer Q given as

[tex]Q=\dfrac{\Delta T}{R_{th}}[/tex]

Where Δt is the temperature difference

[tex]{R_{th}}[/tex] is the thermal resistance.

Connection:

1. Series connection

 [tex]R_{total}={R_1}+{R_2}+{R_1}----{R_n}[/tex]

2. Parallel connection

[tex]\dfrac{1}{R_{total}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}-----\dfrac{1}{R_n}[/tex]

Answer:

Explanation:

CO, carbon monoxide is a toxic gas. It casues asphixiation on people and animals by interfering with hemoglobin, not allowing blood to transport oxygen to the cells in the body.

The normal emissions resulting from the combustion  of fussil fuels are CO2 (carbon dioxide) and H2O (water). Carbon monoxide is formed by an incomplete combustion of fossil fuels or carbon containing fuels in general, this not only produces toxic gas, but also is an inefficient combustion that wastes energy.

Answer:

The acceleration is [tex]2220.00m/s^{2}[/tex]

Explanation:

We know that the acceleration is given by

[tex]a=v\frac{dv}{ds}........................(i)[/tex]

The velocity as a function of position is given by [tex]v=14\cdot s^{7/6}[/tex]

Thus the acceleration as obtained from equation 'i' becomes

[tex]a=14\cdot s^{7/6}\times \frac{d}{ds}(14\cdot s^{7/6})\\\\a=14\cdot s^{7/6}\times \frac{7}{6}\times 14\cdot s^{1/6}\\\\a=\frac{686}{3}\cdot s^{8/6}[/tex]

Hence acceleration at s = 5.5 equals

[tex]a(5.5)=\frac{686}{3}\times (5.5)^{8/6}\\\\\therefore a=2220.00mm/s^{2}[/tex]

Answer: Basically it's a substance with no shape and it's one of the different states of water.

Explanation:

Answer:

Air cooling.

Explanation:

Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.

Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.

Answer:

Fouling :

  When rust or undesired material deposit in the surface of heat exchanger,is called fouling of heat exchanger.

Effect of heat exchanger are as follows"

1.It decreases the heat transfer.

2.It increases the thermal resistance.

3.It decreases the overall heat transfer coefficient.

4.It leads to increase in pressure drop.

5.It increases the possibility of corrosion.

Answer:

The average current will be of 6.36 A.

Explanation:

The anode capacity of magnesium is C = 550 A*h/lb

A month has 30 days.

A day has 24 hours.

Therefore 3 months have:

t = 3 * 30 * 24 = 2160 hours.

The average current is then:

I = C * m / t

I = 550 * 25 / 2160 = 6.36 A

The average current will be of 6.36 A.

Answer:

Height of oil is 7.06 meters.

Explanation:

The situation is shown in the attached figure

The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by

[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(\rho _{water}h_{water}+\rho_{oil}h_{oil})=P_{bottom}-P_{top}[/tex]

Applying the given values we get

[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(1000\times 2.5+800\times h_{oil})=80\times 10^{3}\\\\\therefore 800\times h_{oil}=\frac{80\times 10^{3}}{9.81}-2500\\\\\therefore 800\times h_{oil}=5654.94\\\\\therefore h_{oil}=7.06m[/tex]

Answer:

efficiency =42.62%

AMOUNT OF POWER REJECTED IS 20.080 kW

Explanation:

given data:

power 20 hp

heat energy = 35kW

power production = 20 hp = 20* 746 W = 14920 Watt   [1 hp =746 watt]

[tex]efficiency = \frac{power}{heat\ required}[/tex]

[tex]efficiency = \frac{14920}{35*10^3}[/tex]

                [tex]= 0.4262*10^100[/tex]

                 =42.62%

b) [tex]heat\ rejected = heat\ required - amount\ of\ power\ generated[/tex]

                           [tex]= 35*10^3 - 14920[/tex]

                           = 20.080 kW

AMOUNT OF POWER REJECTED IS 20.080 kW

Answer:

a₁= 1.98 m/s²   : magnitud of the normal acceleration

a₂=0.75  m/s²  : magnitud of the tangential acceleration

Explanation:

Formulas for uniformly accelerated circular motion

a₁=ω²*r : normal acceleration     Formula (1)

a₂=α*r:    normal acceleration     Formula (2)

ωf²=ω₀²+2*α*θ                             Formula (3)

ω : angular velocity

α : angular acceleration

r  : radius

ωf= final angular velocity

ω₀ : initial angular velocity

θ :   angular position theta

r  : radius

Data

r =0.4 m

ω₀= 1 rad/s

α=0.3 *θ , θ= 2π

α=0.3 *2π= 0,6π rad/s²

Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.

We calculate ωf with formula 3:

ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687

ωf=[tex]\sqrt{24.687}[/tex] =4.97 rad/s

a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²    

a₂=α*r = 0,6π * 0.4 = 0.75  m/s²  

Answer:

convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.

Explanation:

convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.

Convection is the transition of heat between distinct temperature fields by moving fluid (liquid or gas). Dry air is less thick than moist air and in the presence of a temperature difference, convection currents can form.

Answer:

final pressure is 6847.41 kPa

Explanation:

given data:

[tex]P_1 =  10,000 kPa[/tex]

[tex]T_1 =520\ degree\ celcius = 793 K[/tex]

[tex]T_2  = 270 degree celcius = 543 K[/tex]

as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream

for ischoric process we know that

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]

[tex]\frac{10*10^6}{793} = \frac{P_2}{543}[/tex]

[tex]P_2 = 6.84741*10^6 Pa[/tex]

final pressure is 6847.41 kPa

Answer:

The displacement is -48m.

Explanation:

Velocity is the rate of change of displacement. So, the instantaneous displacement is calculated by the integration of velocity function with respect to time. Displacement in range of time is calculated by integrating the velocity function with respect to time with in time range.  

Given:

Velocity along the s-axis is

[tex]s{}'=40-3t^{2}[/tex]

time range is t=2s to t=6s.

Calculation:

Step1

Displacement in the time range t=2s to t=6s is calculated as follows:

[tex]\frac{\mathrm{d}s}{\mathrm{d}t}=40-3t^{2}[/tex]

[tex]ds=(40-3t^{2})dt[/tex]

Step2

Integrate the above equation with respect to time with the lower limit as 2 and upper limit as 6 as follows:

[tex]\int ds=\int_{2}^{6}(40-3t^{2})dt[/tex]

[tex]s=(40\times6-40\times2)-3(\frac{1}{3})(6^{3}-2^{3})[/tex]

s=160-208

s=-48m

Thus, the displacement is -48m.

The displacement of a point mass moving along the s-axis from t = 2 s to t = 6 s, with a given velocity function s' = 40 - 3t^2 m/s, is calculated by integrating the velocity to get the position function and then evaluating it between the two time limits, resulting in a displacement of 152 meters.

The question requires finding the displacement of a point mass moving along the s-axis between t = 2 s and t = 6 s. The velocity is given as s' = 40 - 3t^2 m/s. To find the displacement, we will integrate the velocity function with respect to time from 2 to 6 seconds.

To integrate s' = 40 - 3t^2, we get:

The displacement (s) is thus
s = 40t - t^3 evaluated from t = 2 to t = 6. Inserting the limits, we subtract the value at t = 2 from the value at t = 6:

s(6) - s(2) = (40(6) - 6^3) - (40(2) - 2^3) = 240 - 216 - 80 + 8 = 152 m

So, the displacement of the particle from t = 2 s to t = 6 s is 152 meters.

Answer:

a)Humidity ratio =0.013  kg/kg

b) Enthalpy=60.34 KJ/kg

c) Density = 1.16 [tex]Kg/m^3[/tex]

d) Dew point temperature = 64.9°F

e) Wet bulb temperature = 69.53°F

Explanation:

Given that

Dry bulb temperature = 80°F

relative humidity = 60%

Given air is at sea level it means that total pressure will 1 atm.

So P= 1 atm.

As we know that psychrometric charts are always drawn at constant pressure.

Now from charts

We know that dry bulb temperature line is vertical and relative humidity line is curve and at that point these two line will meet ,will us property all property.

a)Humidity ratio =0.013  kg/kg

b) Enthalpy=60.34 KJ/kg

c) Density = 1.16 [tex]Kg/m^3[/tex]

d) Dew point temperature = 64.9°F

e) Wet bulb temperature = 69.53°F

Answer:

T=340. 47 K

Explanation:

Given that

Volume of tank =0.5 [tex]m^3[/tex]

Pressure P=120 KPa

Temperature T=300 K

Added heat ,Q= 20 KJ

Given that air is treated as ideal gas and specific heat is constant.

Here tank is rigid so we can say that it is constant volume system.

We know that specific heat at constant volume for air

[tex]C_v=0.71\ \frac{KJ}{kg.K}[/tex]

We know that for ideal gas

P V = m R T

For air R=0.287 KJ/kg.K

P V = m R T

120 x 0.5 = m x 0.287 x 300

m=0.696 kg

[tex]Q=mC_v\Delta T[/tex]

Lets take final temperature of air is T

Now by putting the values

[tex]Q=mC_v\Delta T[/tex]

[tex]20=0.696\times 0.71\times (T-300)[/tex]

T=340. 47 K

So the final temperature of air will be 340.47 K.

Answer:

maximum length of the specimen before deformation = 200 mm

Explanation:

Hi!

If we have a cylinder with length L₀ , and it is elasticaly deformed ΔL (so the final length is L₀ + ΔL), the strain is defined as:

[tex]\epsilon =\frac{\Delta L}{L_0}[/tex]

And the tensile stress is:

[tex]\sigma = \frac{F}{A}\\F = \text{tensile load}\\A = \text{ cross section area}[/tex]

Elastic modulus E is defined as:

[tex]E = \frac{\sigma}{\epsilon }[/tex]

In this case ΔL = 0.45 mm and we must find maximum L₀. We know that A=π*r², r=(3.3/2) mm. Then:

[tex]\sigma=\frac{2340N}{\pi (1.65 \;mm)^2}=273.68 MPa = [/tex]

[tex]E=123\;GPa=\frac{L_0 \;(273.68MPa)}{0.45\;mm } \\ L_0 = 200\; mm[/tex]

Answer:

Answered

Explanation:

Open- loop 4 bar linkage system:

Open- loop 4 bar linkages are not mechanically constrained, meaning in this type of linkages the degree of freedom is are more than 1.

DOF> 1 example = industrial robots, epicyclic gear trains etc.

Closed loop four bar linkage system:

These types are linkages are mechanically constrained and have degree of freedom as one. examples,  four bar chains,  slider crank mechanism.

DOF=1

Answer:

factor of safety for A36 structural steel is 0.82

Explanation:

given data:

side of column = 3.5 inches

wall thickness = 0.225 inches

load P = 22 kip

Length od column = 9 ft

we know that critical stress is given as

[tex]\sigma_{cr} = \frac{\pi^2 E}{(l/r)^2}[/tex]

where

r is radius of gyration[tex] = \sqrt{\fra{I}{A}}[/tex]

Here I is moment od inertia [tex]= \frac{b_1^2}{12} - \frac{b_2^2}{12}[/tex]

[tex] I == \frac{3.5^2}{12} - \frac{3.05^2}{12} = 5.294 in^4[/tex]

For hollow steel area is given as [tex]A = b_1^2 -b_2^2[/tex]

[tex]A = 3.5^2 -3.05^2 = 2.948 in^2[/tex]

critical stress [tex]\sigma_{cr} = = \frac{\pi^2\times 29\times 10^6}{((9\times12)/(1.34))^2}[/tex]

[tex]\sigma_{cr} =  44061.56 lbs/inc^2[/tex]

considering Structural steel A36

so A36[tex] \sigma_y = 36ksi[/tex]

factor of safety  [tex]= \frac{yield\ stress}{critical\  stress}[/tex]

factor of safety =[tex]\frac{36\times10^3}{44061.56} = 0.82[/tex]

factor of safety for A36 structural steel is 0.82

The student's query involves calculating the factor of safety for a hollow steel column, but without complete material strength data, the calculation cannot be precisely done.

The student's question pertains to determining the factor of safety for a hollow steel column with specified dimensions, material properties (steel E), and loading conditions. To calculate the factor of safety for the column, we would need to compare the column's actual stress under load to its maximum allowable stress. However, due to insufficient data about material yield strength or ultimate strength and considering the provided information does not match the examples given, this calculation cannot be accurately completed without further specifics on the steel's properties.

The resultant force on one side of a 25cm diameter circular plate at the bottom of 3m of pool water is approximately 1447 Newtons, calculated using the principles of fluid pressure and hydrostatic force.

The question asks for the resultant force on one side of a 25cm diameter circular plate at the bottom of 3m of pool water. To calculate this, we first need to understand that the pressure exerted by a fluid in a static situation is given by the equation P = ρgh, where ρ is the density of the fluid (water in this case, which is approximately 1000 kg/m³), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the depth of the fluid above the point of interest (3m in this case).

To find the force exerted by the water on the plate, we also need the area of the plate, which can be derived from its diameter (D = 0.25m). The area (A) of a circle is given by πD²/4, which in this case gives us an area of approximately 0.0491 square meters. The force (F) exerted by the water can then be calculated using the equation F = P*A.

Substituting the values into the equations gives us a pressure (P) of 29430 Pa and, subsequently, a resultant force of approximately 1447 Newtons. This represents the sum of all the hydrostatic forces acting perpendicularly to the surface area of the circular plate due to the water pressure at the depth of 3 meters.

Answer:

maximum tensile load Pmax is 11.91 ksi

Explanation:

given data

area = 0.65 in²

design factor of safety = 3

to find out

what is the maximum tensile load Pmax

solution

we know here area is 0.65 in² and FOS = 3

so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi

so

we use here design factor formula that is

[tex]\frac{ \sigma y}{FOS}  = \frac{Pmax}{area}[/tex]  .............1

put here all these value we get Pmax in equation 1

[tex]\frac{55}{3}  = \frac{Pmax}{0.65}[/tex]

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi

Answer:

0.815 mm

Explanation:

The rod in made of Kevlar 49, so it has an Young's modulus of

E = 125 GPa

The stiffness of a rod is given by:

k = E * A / L

k = E * π/4 * d^2 / L

So:

k = 125*10^9 * π/4 * 0.01^2 / 0.1 = 98.17 MN/m

Of the pulling forces only one is considered because when you pull on something there is always another force on the other side of equal magnitude and opposite direction to maintain equilibrium.

Hooke's law:

Δl = P/k

Δl = 80*10^3 / 98.17*10^6 = 0.000815 m = 0.815 mm

Answer:

Explanation:

Given

Initial Thickness=45 mm

Final thickness=20 mm

Roll diameter=600 mm

Radius(R)=300 mm

coefficient of friction between rolls and strip ([tex]\nu [/tex])=0.15

maximum draft[tex](d_{max})=\nu ^2R[/tex]

[tex]=0.15^2\times 300=6.75 mm[/tex]

Minimum no of passes[tex]=\frac{45-20}{6.75}=3.70\approx 4[/tex]

(b)draft per each pass

[tex]d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}[/tex]

[tex]d=\frac{45-20}{4}=6.25 mm[/tex]