The acceleration due to gravity at sea level is g=9.81 m/s^2. The radius of the earth is 6370 km. The universal gravitational constant is G = 6.67 × 10−11 N-m^2/kg^2. Use this information to determine the mass of the earth.

Answer:

Strain gauge:

 Strain gauge is a sensor. It is use to measure the strain of the material with help of electric resistance. A strain gauge is attached to the work piece and when any change in dimensions of work piece occurs ,due to this electric resistance of strain gauge also changes. Then by using a proper electric circuit strain of a material can be measured. Depending upon the situations one or more than one strain gauges is attached to the work piece to measure the strain .

Use of strain gauges

1. To measure the stress of a structure.

2. Use for testing of ships ,vehicle ,dams etc.

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:

Stress is a force that acts on a unit area of a material. The strength of a material is how much stress it can bear without permanently deforming or breaking.

Is the beam design acceptable for a SF of 2? YES

Explanation:

Your factor of safety is 2, this means your stress allowed is:

Where:

Now we are going to calculate the shear stress and bending stresses of the proposed scenario. If the calculated stresses are less than the allowed stress, that means the design is adequate for a factor of safety of 2.

First off we calculate the reaction force on your beam. And for this you do sum of forces in the Y direction and equal to 0 because your system is in equilibrium:

Knowing this reaction force you can already calculate the shear stress on the cantilever beam:

Now, you do a sum of moments at the fixed end of your cantilever beam, so you can cancel off any bending moment associated with the reaction forces on the fixed end, and again equal to 0 because your system is in equilibrium.

Knowing the maximum bending moment you can now calculate your bending stress as follows:

Where:

Out of the 3 values needed, we already know M. But we still need to figure out c and Ix. Getting c is very straight forward, since you have a rectangle with base (b) 2 and height (h) 5, you know the centroid is right at the center of the rectangle, meaning that the distance from the centroid to the outermost fibre would be 5in/2=2.5in

To calculate the moment of Inertia, you need to use the formula for the second moment of Inertia of a rectangle and knowing that you will use Ix since you are bending over the x axis:

Now you can use this numbers in your bending stress formula:

The shear stress is 10psi and the bending stress is 120psi, this means you are way below the stress allowed which is 50,000 psi, thus the beam design is acceptable. You could actually use a different geometry to optimize your design.

Answer:

c) normal

Explanation:

The pressure forces on a submersed object will be normal to the surfaces of the object. Hydrostatic pressure cannot apply tangential forces, so only the normal components exist.

Also, it should be noted that in the hydrostatic case (submerged object) all pressures depend linearly of depth. For small object we can approximate this with equal pressures everywhere.

Answer:

The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.

Explanation:

The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform.  In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.

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:

froce by 6 gallon liquid on moon surface is  2.86 lbf

Explanation:

given data:

at earth surface

volume of an incompressible liquid = Ve = 25 gallons

force by liquid = 70 lbf

on moon

volume of  liquid = Vm = 6 gallons

gravitational acceleration on moon is am = 5.51 ft/s2

Due to incompressibility , the density remain constant.

mass of liquid on surface of earth[tex]= \frac{ force}{ acceleration}[/tex]

[tex]mass = \frac{70lbf}{32.2 ft/s2}[/tex]

mass = 2.173 pound

[tex]density \rho = \frac{mass}{volume}[/tex]

                  [tex]= \frac{2.173}{25} = 0.0869 pound/ gallon[/tex]

froce by 6 gallon liquid on moon surface is

Fm = mass * acceleration

      = density* volume * am

      = 0.0869 *6* 5.51

      = 2.86 lbf

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

Explanation:

Answer:

Explanation:

Ductile materials typically have a higher ultimate strength because they stretch absorbing more energy before breaking. While fragile materials snap in half before larger deformations due to larger loads occur.

It should be noted that when ductile materials stretch their section becomes smaller, and in that reduced section the stresses concentrate.

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:

[tex]\eta =\dfrac{1}{mL}[/tex]

[tex]R=\sqrt {\dfrac{KA}{hP}}[/tex]

Explanation:

Fin efficiency:

 Fin efficiency is the ratio of actual heat transfer through fin to the maximum heat transfer through fin.

For infinite long fin:

 Actual heat transfer

[tex]q=\sqrt{hPKA}\Delta[/tex]

So the efficiency of fin

[tex]\eta =\dfrac{1}{mL}[/tex]

Where

[tex]m =\sqrt{\dfrac{hP}{KA}}[/tex]

h is heat transfer coefficient,P is the perimeter,K is the thermal conductivity and A is the cross sectional area.

[tex]q=\sqrt{hPKA}\Delta[/tex]\

From equation we can say that fin resistance

[tex]R=\sqrt {\dfrac{KA}{hP}}[/tex]

Thermal resistance offer resistance to flow of heat.

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:

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:

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:

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:

mass of LNG: 129501.3388 kg

quality: 0.005048662

Explanation:

Volume occupied by liquid:

400 m^3*0.9 = 360 m^3

Volume occupied by vapor

400 m^3*0.1 = 40 m^3  

A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present

For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg

From specific volume definition:

vf = liquid volume/liquid mass

liquid mass = liquid volume/vf

liquid mass = 360 m^3/0.002794 m^3/kg

liquid mass = 128847.5304 kg

vg = vapor volume/vapor mass

vapor mass = liquid volume/vg

vapor mass = 40 m^3/0.06118 m^3/kg

vapor mass = 653.8084341 kg

total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg

Quality is defined as the ratio between vapor mass and total mass

quality =  653.8084341 kg/129501.3388 kg = 0.005048662

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

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.

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:

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:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

or using the diameter:

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:

The tube diameter is 2.71 mm.

Explanation:

Given:

Open glass tube is inserted into a pan of fresh water at 20°C.

Height of capillary raise is four times tube diameter.

h = 4d

Assumption:

Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.

Take surface tension of water at 20°C as [tex]72.53\times 10^{-3}[/tex] N/m.

Take density of water as 100 kg/m3.

Calculation:

Step1

Expression for height of capillary rise is gives as follows:

[tex]h=\frac{4\sigma\cos\theta}{dg\rho}[/tex]

Step2

Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:

[tex]4d=\frac{4\times72.53\times10^{-3}\cos0^{\circ}}{d\times9.81\times1000}[/tex]

[tex]d^{2}=7.39\times10^{-6}[/tex]

[tex]d=2.719\times10^{-3}[/tex] m.

Or

[tex]d=(2.719\times10^{-3}m)(\frac{1000mm}{1m})[/tex]

d=2.719 mm

Thus, the tube diameter is 2.719 mm.

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:

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.

The final gage pressure of the gas after being heated from 27°C to 77°C in a closed, rigid tank can be found using Gay-Lussac's Law. After adjusting the initial and final temperatures to Kelvin and converting the gage pressure to absolute pressure, calculate the new absolute pressure and then convert back to gage pressure.

To determine the final gage pressure of a gas modeled as an ideal gas in a closed, rigid tank after it is heated from an initial temperature of 27°C to a final temperature of 77°C, we can use the ideal gas law in a form that relates pressure and temperature while keeping the volume and number of moles constant (Gay-Lussac's Law). We're told that the initial gage pressure is 300 kPa, and we need to find the final gage pressure. The local atmospheric pressure is given as 1 atm, which is equivalent to approximately 101.325 kPa. The formula for the final pressure, assuming no changes in the amount of gas or volume, is P2 = P1 * (T2/T1), where pressures are absolute pressures.

The steps to solve the problem are:

The answer explains how to calculate the initial pressure of a gas using the combined gas law equation.

The initial pressure of the gas in the closed system can be found using the combined gas law equation:

P1V1/T1 = P2V2/T2

Substitute the given values to find the initial pressure, which is calculated to be 1.6 atm.

Therefore, the initial pressure of the gas was 1.6 atm.