We discover a nearby star with two planets. The first planet has an orbit period of 10 years and is in a circular orbit with radius 106 km. The second planet has an orbit period of 15 years. What is its orbit radius? You may assume it is also in a circular orbit.

Answers

Answer 1

Answer: 139 Km.

Explanation:

The question tells us that a planet A has an orbit period of 10 years and its circular orbit has a radius of 106 Km, whilst a planet B has an orbit period of 15 years (also assuming a circular orbit), both orbiting a nearby star.

This information allow us to use the Kepler's 3rd law, for the special case in which the orbit is circular.

Kepler's 3rd law, tells that there exist a direct proportionality between the square of the orbit period, and the cube of the orbit radius (in the more general case, with the cube of the semi-major axis of the elipse), for celestial bodies orbiting a same star.

(like Earth and Mars orbiting Sun).

So, for planet A and planet B (orbiting a same star), we can write the following:

(TA)²/ (TB)² = (rA)³ / (rB)³

Replacing by TA= 10 years, TB= 15 years, rA= 106 Km, and solving for Rb, we get  RB= 139 Km.


Related Questions

What is the governing ratio for thin walled cylinders?

Answers

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:

if [tex]\frac{radius}{thickness} >10[/tex] then you have a thin walled cylinder

or using the diameter:

if [tex]\frac{diameter}{thickness} >20[/tex] then you have a thin walled cylinder

A U tube manometer has water poured into the left side and oil (density is 790 kg/m^3) poured into to the right side. The water in left side is measured to be 70 cm high, while the right side contains both oil and water with oil height 4 times as high as water height. Determine the height of oil and height of water in right side of U tube.

Answers

Answer:

1) Height of oil in right limb = 67.31 cm

2) Height of water in the right limb = 16.83 cm

Explanation:

The U-tube manometer is shown in the attached figure

Foe equilibrium The pressure at the bottom of the U tube should be same

Let the height of the water in the left limb of the manometer be [tex]h_L[/tex]

Thus the pressure at the bottom is found using the equation of pressure statics as

[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{L}.............(i)[/tex]

Similarly for the liquid in the right limb the pressure at the bottom is the sum of the oil column and the water column

Thus we can write

[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}...........(ii)[/tex]

Equating the equations 'i' and 'ii' we get

[tex]P_{atm}+\rho _{water}\times g\times h_{L}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}\\\\\rho _{water}\times g\times 0.7=\rho _{oil}\times 4h\times g+\rho _{water}\times h\times g\\\\\therefore h_{}=\frac{0.7\times \rho _{water}}{4\times \rho _{oil}+\rho _{water}}\\\\h_{}=\frac{0.7\times 1000}{4\times 790+1000}=16.83cm[/tex]

Thus the height of oil is [tex]4\times 16.83=67.31cm[/tex] andthe height of water in the right limb is 16.83 cm.

Describe the differences, if any, between fin efficiency and fin resistance.

Answers

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.

25 gallons of an incompressible liquid exert a force of 70 lbf at the earth’s surface. What force in lbf would 6 gallons of this liquid exert on the surface of the moon? The gravitational acceleration on the surface of the moon is 5.51 ft/s2.

Answers

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

One way to lower NOx emissions in diesel engines is to add water to the fuel. Since water is not mixable with diesel fuel, it has to been emulsified by a surfactant. Explain how it works and why.

Answers

After the emulsion is made, fuel droplet containing water droplets increase its temperature inside the engine,  because of that, water droplets explode causing water rapid evaporation that break down fuel droplet, in consequence, smaller fuel droplets are generated. This makes the temperature of the flame during the combustion decrease, and that is why the reduction of NOx is accomplished by this method .

The 10mm diameter rod is made of Kevlar 49. Determine the change in
length and the change in diameter.
Lenght of the rod = 100mm
Therefore there are two 80KN forces are pulling the rod from both
sides.

Answers

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

The pressure forces on a submersed object will be (A)- Tangential to the objects body (B)- Parallel (C)- Normal (D)- None of the above

Answers

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.

Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a pressure of 101 kPa on top of the 0.5-m-diameter piston. The gage pressure of the air inside the cylinder is 1.2 kPa. The local acceleration of gravity is g = 9.81 m/s2 . Subsequently, a weight is placed on top of the piston causing the piston to fall until reaching a new static equilibrium position. At this position, the gage pressure of the air inside the cylinder is 2.8 kPa. Determine (a) the mass of the piston, in kg, and (b) the mass of the added weight, in kg

Answers

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

Which of the two materials (brittle vs. ductile) usually obtains the highest ultimate strength and why?

Answers

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.

Consider a nuclear power plant that produces 1200 MW of power and has a conversion efficiency of 34 percent (that is, for each unit of fuel energy used, the plant produces 0.34 units of electrical energy. Assuming continuous operation, determine the amount of nuclear fuel consumed by the plant per year.

Answers

Answer with Explanation:

The relation between power and energy is

[tex]Energy=Power\times Time[/tex]

Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals

[tex]E=1200\times 10^{6}\times 3600\times 24\times 365=3.784\times 10^{16}[/tex]

Now from the energy mass equivalence we have

[tex]E=mass\times c^2[/tex]

where

'c' is the speed of light in free space

Thus equating both the above values we get

[tex]3.784\times 10^{16}=mass\times (3\times 10^{8})^{2}\\\\\therefore mass=\frac{3.784\times 10^{16}}{9\times 10^{16}}=0.42kg[/tex]

Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is

[tex]mass_{req}=\frac{mass}{\eta }=\frac{0.43}{0.34}=1.235[/tex]

Thus 1.235 kg of nuclear fuel is reuired for operation.

Tensile strength is a measure of a material’s resistance to being pulled apart. a) True b) False

Answers

Answer:

The given statement is correct.

Explanation:

When we load a material axially stresses are developed in the material. these stresses arise internally to keep the object in equilibrium.

When a tensile load is applied to an material since the nature of the force is to cause elongation in the material, axial strain is developed in the material or we can say that any arbitrary point in the material undergoes a displacement in the direction of the applied load. The stresses that are developed in the material tend to tear the material apart.

Strength of a material by definition is it's ability to resist deformation, thus tensile strength can be defined as the resistance of the material to tensile strains which have the tendency to tear the material apart.

convert
a) 760 miles/hour to meters/second
b) 921 kg/cubic meter to pound mass/cubic foot
c) 5.37 x 10^3 kJ/ min to hp.

Answers

Answer:

(a)[tex]1.308\times 10^{-4}m/sec[/tex]

(b)57.33831 pound/cubic feet

(c)120.1095 hp

Explanation:

We have

(a) 760 miles / hour

We know that [tex]1\ mile\ =0.00062m[/tex]

And 1 hour = 60×60=3600 sec

So [tex]760miles/hour=\frac{760\times 0.00062meter}{3600sec}=1.308\times 10^{-4}m/sec[/tex]

(b) 927 kg/cubic meter to mass/cubic foot

We know that 1 kg = 2.20 pound

So 921 kg = 921×2.20=2026.2 pound

We know that 1 cubic meter = 35.31 cubic feet

So 57.33831 pound / cubic feet

(c) We have to convert to hp

[tex]5.37\times 10^3kj/min=\frac{5.37\times 1000kj}{60sec}=89.5kj/sec[/tex]

We know that 1 kj /sec = 1.341 hp

So 89.5 kj/sec = 89.5××1.341=120.1095 hp

What is 1000J in Btu?

Answers

Answer:

0.948 Btu

Explanation:

1 Btu = 1055 J so [tex]\frac{1000}{1055}[/tex] = 0.948 Btu

1 BTU = 1055.06 joules **

(1000 J) x (1 BTU / 1055.06 J) = 0.9478 BTU

** Note:  You'll have to accept my conversion factor, since I went and looked it up and you didn't.

A heat engine receives heat from a heat source at 1453 C and has a thermal efficiency of 43 percent. The heat engine does maximum work equal to 539 kJ. Determine: a) the heat supplied to the heat engine by the heat source (kJ), b) the heat rejected to the heat sink (kJ), and c) the temperature of the heat sink (C).

Answers

Answer:

a) 1253 kJ

b) 714 kJ

c) 946 C

Explanation:

The thermal efficiency is given by this equation

η = L/Q1

Where

η: thermal efficiency

L: useful work

Q1: heat taken from the heat source

Rearranging:

Q1 = L/η

Replacing

Q1 = 539 / 0.43 = 1253 kJ

The first law of thermodynamics states that:

Q = L + ΔU

For a machine working in cycles ΔU is zero between homologous parts of the cycle.

Also we must remember that we count heat entering the system as positiv and heat leaving as negative.

We split the heat on the part that enters and the part that leaves.

Q1 + Q2 = L + 0

Q2 = L - Q1

Q2 = 539 - 1253 = -714 kJ

TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:

η = 1 - T2/T1

T2/T1 = 1 - η

T2 = (1 - η) * T1

The temperatures must be given in absolute scale (1453 C = 1180 K)

T2 = (1 - 0.43) * 1180 = 673 K

673 K = 946 C

The primary heat transfer mechanism that sears my skin if I fall in a campfire is: a)- Syncrotonization b)-Radiation c)- Conduction d)-Convection

Answers

Answer:

c)Conduction

Explanation:

As we know that conduction heat transfer is take place due to movement of electron.When we fall on the campfire then the primary heat transfer mechanism is conduction because .

Convection heat transfer taker place due to motion of fluid.

Radiation heat transfer dominates at very high temperature but in campfire the temperature is not too high.

So option C is correct.

Define factor of safety and its significance

Answers

Answer:

Factor of safety is defines as the ratio between the strength of a material and the maximum stress that is developed in the material as a result of any given loading.

Mathematically we can write

[tex]F.O.S=\frac{Strength}{Stress_{working}}[/tex]

Consider any member of a machine that is under state of stress due to a general external loading, now we know that the material has a definite value of strength meaning that there is a definite value of force/stress at which the material will fail by fracture or by yielding as an example say a rod of steel will fail if we apply a load of 1000 Newton's on it.

Since as a basic principle of design we do not want any chance of failure in the machine at any cost we limit the maximum value of the stress that is developed in the machine part by designing the part in such a way that during the application of maximum load in the machine part the maximum stresses that are developed in the machine are well below the strength of the material for safety considerations. The factor by which the working stresses are less than the strength is termed as factor of safety.

Another significance of factor of safety is that if we need to estimate the loads on the machine that may act on it and we are not absolutely sure about the magnitude of the loads then we tend to increase the loads by a factor so that we design the structure to withstand greater loads hence if in case in the lifetime of the structure the loads exceed the normal value our structure will still remain structurally safe.  

The kinetic energy correction factor depends on the (shape — volume - mass) of the cross section Of the pipe and the (velocity — pressure — temperature distribution.

Answers

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.

A cylindrical specimen of some metal alloy having an elastic modulus of 123 GPa and an original cross-sectional diameter of 3.3 mm will experience only elastic deformation when a tensile load of 2340 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.

Answers

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]

Briefly describe the operation and use of strain gauges

Answers

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.

A component is composed of 0.0633 moles of platinum and 0.044 moles of nickel. Determine (a) total mass of the component (b) weight % of platinum.

Answers

Answer:

(a) Total mass = 14.931 kg

(b) Weight % of platinum = 17.29 %

Explanation:

(a) We have given moles of platinum = 0.0633 moles

Molar mass of platinum = 195.084 unit

Now mass of platinum = number of moles × molar mass = 0.0633×195.084 = 12.3488 gram

We have given moles of nickle = 0.044

Molar mass of nickle = 58.6934 unit

So mass of nickle = 0.044×59.6394 =2.5825 gram

So total mass = 12.3488+2.5825 = [tex]14.931[/tex]

(b) Weight % of platinum = [tex]\frac{weight\ of\ platinum}{total\ weight}=\frac{2.5825}{14.93}=0.1729[/tex] = 17.29 %

What system of units is superior, SI, BG, or EE? Why?

Answers

Answer:

SI unit of system is  superior.

Explanation:

The metric system is used internationally for measurement of unit.This metric system of unit is easier to understand .

Metric system of unit also known as international system of unit.The international system of unit is also represent in short form like SI unit.This unit system is internationally accepted.

So SI unit of system is  superior.

A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 90% liquid and 10% vapor, by volume, at 150 K, what mass of LNG (kg) will the tank hold? What is the quality in the tank?

Answers

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

The bars of the truss each have a cross-sectional area of 1.25 in2. If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss.

Answers

Answer:

P=25000lbf

Explanation:

For this problem we will use the equation that relates, the effort, the area and the force for an element under normal stress.

σ=P/A

σ=stress=20kSI=20 000 lbf/in ^2

P=force

A=area

solving for P

P=Aσ

P=(20 000 lbf/in ^2)(1.25in^2)

P=25000lbf

Answer:

The maximum magnitude P of the loads that can be applied to the truss = 25,000 Pounds or 111,205.5 Newtons.

Explanation:

In order to calculate the maximum load P, we will make use of the formula: Maximum average stress (20 ksi) = maximum load P ÷ cross-sectional area (1.25 in²)

Make P (the maximum load) the subject of the formula: P = 20 ksi × 1.25 in².

Before moving further, we have to convert the average normal stress (in ksi) to an appropriate unit: The average normal stress = 20 ksi = 20 kip per square inch (kip/in²)

But 1 kip = 1000 Pounds (i.e., 1000 lb)

Therefore, 20 ksi = 20,000 Pounds/in².

Therefore, P (maximum load) = 20,000 pounds/in² × 1.25 in² = 25,000 Pounds = 111,205.5 Newtons (because 1 Pound = 4.44822 Newtons).

An open glass tube is inserted into a pan of fresh water at 20 °C. What tube diameter is needed to make the height of capillary rise equal to four times the tube diameter? State all assumptions.

Answers

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.

 

The sticker inside the door of my car says that the tire pressure should be 32 psig (322 kPa) when the tire is cold. Before a road trip, I fill the tire to this pressure on a cold morning when the temperature is 15 °C, and then head out towards Las Vegas. When I make a rest stop in Barstow, it is now quite warm out, and the air in my tires has also warmed up from friction during the long drive. So, the air in the tires is now 60 °C. Assuming my tires don't leak or expand (volume is constant), what is the expected tire pressure at this rest stop?

Answers

Answer:

37 psi

Explanation:

For ideal gases this equation applies:

p1*V1/T1 = p2*V2/T2

Since we are assuming volume remains constant:

V2 = V1

p1/T1 = p2/T2

p2 = p1*T2/T1

The temperatures must be in absolute scale.

T1 = 15 + 273  = 288 K

T2 = 60 + 273 = 333 K

Then:

p2 = 32 * 333 / 288 = 37 psi

Which two forms of energy are usually converted directly from solar radiation for houses in temperate climates. a. Thermal and Electrical b. Chemical and Nuclear c. Mechanical and Electrical

Answers

Answer:

The correct answer is option 'a': Thermal and Electrical energy

Explanation:

We know that the solar radiation that we receive from sun provides us heat. Thus the solar radiation is a natural source of thermal energy which can be utilized in solar cooker's to cook food or to warm water to generate steam in Solar thermal power plants. The steam generated is used to drive a turbine and hence produce electricity.

As we know that solar radiation is converted into directly usable forms of energy by the solar panels. The solar panels work on the principle of photo-electric effect in which light energy is directly converted into electrical energy to run our electrical devices at home such as light bulbs, fans, e.t.c. An excellent application of this principle is the international space station that orbits the earth and is fully powered by solar energy.

The flow of a real fluid has (more —less - same ) complexity of an ideal fluid, owing to the phenomena caused by the existence of (viscosity-pressure drop- friction

Answers

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.

Define the stress and strength? A material has yield strength 100 kpsi. A cantilever beam has length 10 in and a load of 100 Lbf is applied at the free end. The beam cross section is rectangular 2""x5’. Is the beam design acceptable or not for a factor of safety 2?

Answers

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:

σall = YS/FS = 100kpsi/2 = 50kpsi

Where:

σall => Stress allowedYS => Yield StrengthFS => Factor of safety

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:

ΣFy = 0-100 + Ry = 0     thus,Ry = 100 lbf

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

τ = F/Aτ = 100lbf/(2in*5in) τ = 10 psi

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.

ΣM = 0-100lbf*10in + M = 0M = 1000 lbf-in

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

σ = M*c/Ix

Where:

σ => Bending StressM => Bending Momentc => Distance from the centroid of your beam geometry to the outermost fiber.Ix => Second moment area of inertia

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:

Ix = (b*h^3)/12 = (2in*5in^3)/12 = 20.83 in4

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

σ = M*c/Ixσ = 1000 lbf-in * 2.5in / 20.83 in4σ = 120 psi

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.

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas is heated to 77°C, determine the final pressure, expressed as a gage pressure, in kPa. The local atmospheric pressure is 1 atm.

Answers

Final answer:

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.

Explanation:

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:

Convert the initial and final temperatures from Celsius to Kelvin by adding 273.15 to each.Add the atmospheric pressure to the initial gage pressure to get the initial absolute pressure.Use the equation to calculate the final absolute pressure.Subtract the atmospheric pressure from the final absolute pressure to get the final gage pressure.

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.

Which of the following statements is true about machines? a)- Machines are structures made of one or more movable parts. b)- Machines have a least one multiforce member as part. c)- Machines transmit forces or alter the effect . d)- Machines, unlike frames, can’t be constrained

Answers

Answer:

Option c is True

Explanation:

a)- Machines are structures made of one or more movable parts. is false statement because Machines can be made without any movable part as in simple machines like inclined plains, wedges, screw etc.

b) Machines have a least one multi force member as part. is false statement since it is not a compulsory condition.

c) Machines transmit force or alter the effect. This is a true statement as machine are made in  the first place to transmit forces and alter the effect.  

d) Machines can be Contrained.

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