A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-position and 2.5m/s when 225mm from mid-position. Find : i) It's max amplitude ii) Max acceleration iii) The periodic time iv) The frequency of oscillation.

Answer with Explanation:

From newton's second law the acceleration produced by a force on a mass 'm' is given by

[tex]Acceleration=\frac{Force}{Mass}[/tex]

Applying the given values in the above equation we get

[tex]Acceleration=\frac{kvx}{m}[/tex]

Also we know that acceelration of a particle can ve mathem,atically written as

[tex]a=\frac{v\cdot dv}{dx}[/tex]

Applying the given values in the above equation we get

[tex]\frac{kvx}{m}=\frac{v\cdot dv}{dx}\\\\\Rightarrow {kx}\cdot dx=m\cdot dv\\\\\int kxdx=\int mdv\\\\\frac{kx^2}{2}=mv-c[/tex]

'c' is the constant of integration

whose value is found that at x =0 v= [tex]v_o[/tex]

Thus

[tex]c={mv_o}[/tex]

Thus the velocity as a function of position is

[tex]v=\frac{1}{m}(\frac{kx^2}{2}+c)[/tex]

Now  by definition of velocity we have

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

Using the function of velocity in the above relation we get

[tex]\frac{dx}{kx^{2}+\sqrt{2c}}=\frac{dt}{2m}\\\\\int \frac{dx}{(\sqrt{k})^2x^{2}+(\sqrt{2c})^2}=\int \frac{dt}{2m}\\\\\frac{1}{\sqrt{2kc}}\cdot tan^{-1}(\frac{(\sqrt{k})x}{\sqrt{2c}})=\frac{t}{2m}+\phi \\\\[/tex]

where

[tex]\phi [/tex] is constant of integration

Now it is given that at t = 0 ,x = 0

thus from the above equation of position and time we get [tex]\phi =0[/tex]

Thus the position as a function of time is

[tex]x(t)=\sqrt{\frac{2c}{k}}\cdot tan(\frac{kct}{\sqrt{2}m})[/tex]

where c=[tex]mv_o[/tex]

Answer:

The specific volume of the super heated steam is 0.2139 m³/kg.

Explanation:

Super heated steam is the condition where only compressed vapor of water present. This is not containing any liquid form of water. Super heated steam has very high pressure depending upon temperature. If heat supplied increases after saturation vapor condition of the water, the water fuses into steam completely. Properties of super heated steam are taken from the super heated steam table.

Given:  

Temperature of superheated steam is 300°C.  

Pressure of the super heated steam is 1.2 Mpa.

Calculation:

Step1  

Value of specific volume of the super heated steam is taken from superheated steam table at 300°C and 1.2 Mpa as follows:

Specific volume is 0.2139 m³/kg.

Step2

All other properties are also taken from the table as:

Internal energy is 2789.7 kj/kg.

Enthalpy is 3046.3 kj/kg.

Entropy is 7.034 kj/kgK.

The part of table at the temperature 300°C and pressure 1.2 Mpa is shown below:

 Thus, the specific volume of the super heated steam is 0.2139 m³/kg.

Answer:

The weight of the station becomes [tex]3.756\times 10^{6}N[/tex]

Explanation:

Since the acceleration due to gravity decreases with increase in height we conclude that at a height of 350 kilometers the weight of the material will be lesser.

At the ground we have

[tex]W=mass\times g_{surface}\\\\\therefore mass=\frac{W}{g_{surface}}\\\\mass=\frac{4.22\times 10^{6}N}{9.81}\\\\\therefore mass=430173.292kg[/tex]

Now we know that the variation of acceleration due to gravity with height above surface of earth is given by

[tex]g(h)=g_{surface}(1-\frac{2h}{R})[/tex]

where R = 6371 km is Radius of earth

Applying values we get the value of 'g' at height of 350 kilometers equals

[tex]g(350)=9.81\times (1-\frac{2\times 350}{6371})=8.732ms^{-2}[/tex]

hence the weight in orbit becomes

[tex]W_{orbit}=mass\times g_{orbit}\\\\W_{orbit}=430173.292\times 8.732\\\\ \therefore W_{orbit}=3.756\times 10^{6}N\\[/tex]

Answer:

a)True

Explanation:

While machining of ductile material and high feed and low cutting speed welding action take place between tool material and chip material this welding action is called built up edge. Built up edge action takes place due to high temperature.

To decrease the built up edge action

1. Increase the rake angle.

2. Increase the cutting speed

3. Decrease the feed rate

4. Use cutting fluid

The magnitude and direction of the resultant force from two applied forces, P and Q, at a point can be determined graphically using the triangle rule by forming a vector triangle, measuring the resultant vector, and calculating its angle.

To determine the magnitude and direction of the resultant force using the triangle rule when two forces, P and Q, are applied at a point, you can follow a graphical method. Picture two vectors representing these forces originating from the same point. Since P = 60 lb and Q = 25 lb, you draw them to scale, with the tail of Q starting at the head of P to form a triangle. By connecting the tail of P to the head of Q, you form the resultant vector.

To find the magnitude of the resultant force, measure the length of the resultant vector using the same scale and apply trigonometry or the Pythagorean theorem if the angle is known. The direction of the resultant force is given by the angle it makes with either P or Q, which can be measured with a protractor in reference to a baseline, such as the horizontal.

Answer:

The power output of the turbine is 603 KW.

Explanation:

Turbine is the thermodynamic open system in which fluid looses thermal energy into kinetic energy. Kinetic energy then converted into electric energy.

Here, fluid is air which passes through turbine at 800 Kpa and 870 K with a velocity of 60 m/s.

The turbine is an adiabatic turbine that means there is no heat transfer from the surrounding. Finally the air leaves the turbine at 120 Kpa and 520 K with a velocity of 100 m/s. The turbine inlet area is 90 cm2

Given:

Inlet pressure is  [tex]P_{1}=800[/tex]kpa.

Inlet temperature is [tex]T_{1}=870[/tex]K.

Inlet velocity is[tex]V_{1}=60[/tex] m/s.

Outlet pressure is [tex]P_{2}=120[/tex]Kpa.

Outlet temperature is [tex]T_{2}=520[/tex]K.

Outlet velocity is [tex]V_{2}=100[/tex] m/s.

Inlet area of turbine is A=90 cm2.

Step1

Convert the area into SI unit as follows:

[tex]A=90 cm^{2}(\frac{1 m^{2}}{10^{4}cm^{2}})[/tex]

[tex]A=0.009 m^{2}[/tex]

Step 2

Consider air as an ideal gas. So, ideal gas equation is applicable. For air, gas constant is 287 j/kgK.

Ideal gas equation is expressed as follows:

[tex]P=\rho RT[/tex]

Here, P is pressure, T is temperature and \rho is density.

Density of air is calculated by ideal gas equation as follows:

[tex]\rho =\frac{P}{RT}[/tex]

[tex]\rho =\frac{800\times 10^{3}}{287\times870}[/tex]

[tex]\rho =3.2039 kg/m^{3}[/tex]

Step 3

Mass flow rate is calculated as follows:

[tex]\dot{m}=\rho  AV_{1}[/tex]

[tex]\dot{m}=3.2039\times 0.009\times60[/tex]

[tex]\dot{m}=1.73 Kg/s[/tex]

Step 4

Steady state equation is the equation of first law of thermodynamics for the open system

Steady state equation for the turbine as follows:

[tex]h_{1}+\frac{v^{2}_{1}}{2000}+Z_{1}+Q=h_{2}+\frac{v^{2}_{2}}{2000}+Z_{2}+W[/tex]

Heat transfer is zero as the process is adiabatic. So value of Q is zero.

Turbine is taken as at the same level. So the value of  [tex]Z_{1}[/tex] is equal to [tex]Z_{2}[/tex].

Substitute the value of Q as zero and tex]Z_{1}[/tex] is equal to [tex]Z_{2}[/tex] in steady state equation as follows:

[tex]h_{1}+\frac{v^{2}_{1}}{2}+Z_{1}+0=h_{2}+\frac{v^{2}_{2}}{2}+Z_{1}+W[/tex]

[tex]h_{1}+\frac{v^{2}_{1}}{2}+0=h_{2}+\frac{v^{2}_{2}}{2}+W[/tex]

[tex]W=(h_{1}-h_{2})+\frac{v^{2}_{1}-v^{2}_{2}}{2000}[/tex]

[tex]W=c_{p}(T_{1}-T_{2})+\frac{60^{2}-100^{2}}{2000}[/tex]

Specific heat at constant pressure is 1.005 kj/kgK for air.

Substitute the values of temperature and specific heat at constant temperature in the above simplified steady state equation as follows:

W=1.005(870-520)-3.2

W=351.75-3.2

W=348.55 Kj/kg.

Step 5

Power of the turbine is calculated as follows:

[tex]P=\dot{m}W[/tex]

[tex]P=1.73\times348.55[/tex]

P=603 KW

Thus, the power output of the turbine is 603 KW.

In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.  

Answer:

Network Topology

Explanation:

A network topology is the arrangement of nodes usually switches or routers, and connections in a network, often represented as a graph.  The topology of the network, and the relative locations of the source and destination of traffic flows on the network, determine the optimum path for each flow and the extent to which redundant options for routing exist in the event of a failure.

There are two types of network topologies: physical and logical. Physical topology emphasizes the physical layout of the connected devices and nodes, while the logical topology focuses on the pattern of data transfer between network nodes.

I leave you an example in the next picture:

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

[tex]\sigma=\frac{\textup{PD}}{\textup{2t}}[/tex]

where, t is the thickness

on substituting the respective values, we get

[tex]100=\frac{\textup{2\times800}}{\textup{2t}}[/tex]

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

Calculate the elongation of a cylindrical nickel wire under a specific load by using the formula for elastic deformation and provided values.

To calculate the elongation of the cylindrical nickel wire under a load, we can use the formula for elastic deformation: elongation = (F * L) / (A * E), where F is the load, L is the length, A is the cross-sectional area, and E is the Young's modulus.

Substitute the given values: diameter = 1.8 mm (radius = 0.9 mm), load = 290 N, length = 2.6 × 10^4 mm, and Young's modulus for nickel = 207 × 10^9 N/m^2. Solve for the elongation to find the answer.

The elongation of the cylindrical nickel wire under the given load is found to be **insert answer here**.

The elongation of the nickel wire when a load of 290 N is applied is approximately 14.295 meters.

Step 1

To calculate the elongation of the nickel wire under the applied load, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] of a material is directly proportional to the applied force ([tex]\( F \)[/tex]) and the material's elastic modulus ([tex]\( E \)[/tex]), and inversely proportional to its cross-sectional area ([tex]\( A \)[/tex]) and original length ([tex]\( L_0 \)[/tex]). Mathematically, it's expressed as:

[tex]\[ \Delta L = \frac{F \cdot L_0}{A \cdot E} \][/tex]

Where:

- F is the applied force (290 N)

- [tex]\( L_0 \)[/tex] is the original length of the wire (2.6 × [tex]10^4[/tex] mm = 26,000 mm)

- A is the cross-sectional area of the wire

- E is the elastic modulus of nickel (207 × [tex]10^9[/tex] [tex]N/m^2[/tex])

Step 2

First, let's calculate the cross-sectional area (A) of the wire using its diameter (d ):

[tex]\[ A = \frac{\pi d^2}{4} \][/tex]

Given that the diameter [tex](\( d \))[/tex] is 1.8 mm, we have:

[tex]\[ A = \frac{\pi \times (1.8 \times 10^{-3})^2}{4} \][/tex]

Now, let's calculate the elongation ([tex]\( \Delta L \)[/tex]) using Hooke's Law:

[tex]\[ \Delta L = \frac{290 \times 26,000}{A \times 207 \times 10^9} \][/tex]

Step 3

Finally, we can substitute the values and solve for [tex]\( \Delta L \).[/tex] Let's do the calculations.

First, let's calculate the cross-sectional area A:

[tex]\[ A = \frac{\pi \times (1.8 \times 10^{-3})^2}{4} \]\[ A = \frac{\pi \times 3.24 \times 10^{-6}}{4} \]\[ A = 2.55 \times 10^{-6} \, \text{m}^2 \][/tex]

Step 4

Now, let's calculate the elongation [tex](\( \Delta L \))[/tex] using Hooke's Law:

[tex]\[ \Delta L = \frac{290 \times 26,000}{2.55 \times 10^{-6} \times 207 \times 10^9} \]\[ \Delta L = \frac{7,540,000}{528.15} \]\[ \Delta L = 14.295 \, \text{m} \][/tex]

So, the elongation of the nickel wire when a load of 290 N is applied is approximately 14.295 meters.

Answer:

30.25 in Hg will be equal to 14.855 psi

Explanation:

We have given 30.25 Hg pressure

We have to convert the pressure of 30.25 Hg into psi

We know that 1 inch of Hg = 0.4911 psi

So to convert 30.25 inch Hg in psi we have to multiply with 0.4911

We have to convert 30.25 in Hg

So [tex]30.25inHg=30.25\times 0.4911=14.855775psi[/tex]

So 30.25 in Hg will be equal to 14.855 psi

Answer:

Time period will be 1.26 sec

Explanation:

We have given amplitude A = 0.1 M

Speed [tex]\frac{dx}{dt}=0.5m/sec[/tex]

The displacement equation of simple harmonic motion is given by

[tex]x(t)=Asin\omega t[/tex]

Differentiating both side

[tex]\frac{dx}{dt}=A\omega cos\omega t[/tex]

In question it is given that at t=0, x=0 and [tex]\frac{dx}{dt}=0.5m/sec[/tex]

So [tex]0.5=0.1\omega cos0[/tex]

[tex]\omega =5sec^{-1}[/tex]

Now period of oscillation [tex]T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{5}=1.26sec[/tex]

Answer:

[tex]5.31\frac{kg}{m^3}[/tex]

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to [tex]\frac{m}{M}[/tex], where m is the mass and M the molar mass of the gas, and the density is [tex]\frac{m}{V}[/tex].

For air [tex]M=28.66*10^{-3}\frac{kg}{mol}[/tex] and [tex]\frac{5}{9}R=K[/tex]

So, [tex]598.59 R*\frac{5}{9}=332.55K[/tex]

[tex]pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}[/tex]

Answer:

Both Reynolds and surface roughness

Explanation:

For turbulent flow friction factor is a function of both Reynolds and surface roughness of the pipe.But on the other hand for laminar flow friction factor is a function of only Reynolds number.

Friction factor for turbulent flow:

1.   For smooth pipe

[tex]f=0.0032+\dfrac{0.221}{Re^{0.237}}[/tex]

[tex]5\times 10^4<Re<4\times 10^7[/tex]

2.   For rough pipe

[tex]\dfrac{1}{\sqrt f}=2\ log_{10}\frac{R}{K}+1.74[/tex]

Where R/K is relative roughness

Friction factor for laminar flow:

[tex]f=\dfrac{64}{Re}[/tex]

Answer:

 Rage pressure:

The rage pressure in terms of engineering is the protected shield hard plastic shell and made up of the hard material that is basically used in the protection.

It is the effective material and used in the sewing machine so the pressure developed due to the hard material is known as rage pressure and it has strong elasticity.  

 Absolute pressure:

The pressure which is relative to the perfect vacuum is known as absolute pressure. The absolute pressure is basically measured against the atmospheric pressure.

The absolute pressure is defined as the total pressure in the fluid at the point is equal to the sum of the atmospheric pressure and gauge.  

Answer:

304.13 mph

Explanation:

Data provided in the question :

The Speed of the flying aircraft = 300 mph

Tailwind of the true airspeed = 50 mph

Now,

The ground speed will be calculated as:

ground speed = [tex]\sqrt{300^2+50^2}[/tex]

or

The ground speed = [tex]\sqrt{92500}[/tex]

or

The ground speed = 304.13 mph

Hence, the ground speed is 304.13 mph

Answer:

In Rankine 524.07°R

In kelvin 291 K

In Fahrenheit 64.4°F  

Explanation:

We have given temperature 18°C

We have to convert this into Rankine R

From Celsius to Rankine we know that  [tex]T(R)=(T_{C}+273.15)\frac{9}{5}[/tex]

We have to convert 18°C

So [tex]T(R)=(18+273.15)\frac{9}{5}=524.07^{\circ}R[/tex]

Conversion from Celsius to kelvin

[tex]T(K)=(T_{C}+273)[/tex]

We have to convert 18°C

[tex]T(K)=(18+273)=291K[/tex]

Conversion of Celsius to Fahrenheit

[tex]T(F)=T_{C}\times \frac{9}{5}+32=64.4^{\circ}F[/tex]

Answer:

The problem is that the pumps would consume more energy than the generators would produce.

Explanation:

Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.

A pump uses electricity to add energy to the water to send it to a higher potential energy state.

Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.

What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.

Answer and Explanation:

Young's modulus  is a mechanical property that estimates the solidness of a strong material. If we have information about stress and strain the we can easily found the young's modulus.

Young's modulus gives us information that how hard or how easy to bend a solid material

Young's modulus is given by [tex]young's\ modulus=\frac{stress}{strain}[/tex]

Stress = [tex]\frac{force}{area}[/tex] and strain is [tex]=\frac{chane\ in\ length}{actual\ length}=\frac{\Delta L}{L}[/tex]

Answer:

No, it absolute pressure.

Explanation:

Gauge pressure is relative to the pressure of the atmosphere. It is the difference between the pressure measured and the pressure of the atmosphere. If it is measuring atmospheric pressure it will always read zero.

The measurement is an absolute pressure, which is the pressure above a total vacuum.

Answer with Explanation:

Stress is defined as the force acting per unit area on a material.

Mathematically

[tex]\sigma =\frac{dF}{dA}[/tex]

where

[tex]\sigma [/tex]  is the stress ,[tex]dF[/tex] is an infinitesimal force that acts on an infinitesimal area [tex]dA[/tex]

When a body is under stress it's dimensions change and this change in dimensions is known as strain.

Mathematically

[tex]\epsilon =\frac{\Delta x}{X}[/tex]

where

[tex]\epsilon=[/tex] strain in the object

[tex]\Delta x =[/tex] is the change in any dimension of the body

Now in the above relation of stress, the area involved  also changes when the body is loaded as the load produces strain which changes the dimensions of the body.

Now while calculating the stress if we use the original area of the cross section of the body prior to loading the stress that we calculate is the engineering stress and the strain associated with it is the engineering strain.

On the other hand if we use the true cross section of the body when it is loaded  the stress that we calculate is the true stress and the strain associated with it is the true strain.

Mathematically they are related as

[tex]\epsilon _{true}=ln(1+\epsilon _{engineering})}[/tex]

Thus the true stress is found to be larger than engineering stress.

Answer:

The volume flow rate of air is [tex]Q=A\times V[/tex]

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

[tex]Volume=Area\times V=A\times V[/tex]

Hence the volume flow rate is [tex]Q=V\times A[/tex]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × [tex]10^{-2}[/tex] N/m

so average radius = [tex]\frac{diameter}{2}[/tex] =  100 µm = 100 ×[tex]10^{-6}[/tex] m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = [tex]\frac{2 \sigma }{r}[/tex]    

put here value

Δp = [tex]\frac{2*2.69*10^{-2}}{100*10^{-6}}[/tex]  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

The difference in pressure between the inside and outside of small droplets of carbon tetrachloride at 68 °F with a diameter of 200 um is calculated using the Laplace pressure equation and results in a difference of 2700 dyne/cm2.

The relationship between the pressure inside and outside of small droplets is described by the Laplace pressure equation. The Laplace equation for the pressure difference is ΔP = 2σ/r, where ΔP is the pressure difference, σ is the surface tension, and r is the radius of the droplet. For carbon tetrachloride, the surface tension at 20°C (68°F) is approximately 27 dyne/cm. Therefore, the pressure difference (ΔP) would be 2 x 27 dyne/cm divided by the radius in cm. Given that 200 um = 0.02 cm, the pressure difference is ΔP = 2 x 27 dyne/cm / 0.02 cm = 2700 dyne/cm2. This means that the pressure inside the droplets is 2700 dyne/cm2 greater than the pressure outside the droplets.

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Answer:

B) Process

Explanation:

In thermodynamics a process is a passage of a thermodynamic system from an initial to a final state of thermodynamic equilibrium.

A thermodynamic process path is the series of states through which a system passes from an initial to a final state.

Cycle is a process in which initial and final state are identical.

Answer:

Less material waste and time.

Explanation:

Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.

There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.

Answer:

Density of oil will be 897.292 kg[tex]m^3[/tex]

And specific gravity of oil will be 0.897

Explanation:

We have given density of oil is 1.74 slugs/[tex]ft^3[/tex]

We have to convert this slugs/[tex]ft^3[/tex] into kg/[tex]m^3[/tex]

We know that 1 slugs = 14.5939 kg

So 1.74 slug = 1.74×14.5939 = 25.3933 kg

And 1 cubic feet = 0.0283 cubic meter

So [tex]1.74slug/ft^3=\frac{1.74\times 14.5939kg}{0.0283m^3}=897.292kg/m^3[/tex]

Now we have to calculate specific gravity it is the ratio of density of oil and density of water

We know that density of water = 1000 kg/[tex]m^3[/tex]

So specific gravity of water [tex]=\frac{897.292}{1000}=0.897[/tex]

Answer with Explanation:

For a linearly responsive system principle of superposition states that

"The cumulative response of the given system to forces of different magnitudes is the sum of the individual responses of the system to the individual forces"

In a less formal manner principle of superposition states that the effect of various forces acting together on any body is the sum of all the effects on the body produced when each force acts individually.

Now we know that

[tex]Stress=\frac{Force}{Area}[/tex]

Since stress at any point is in linear relation with the force hence we can conclude that the state of stress at any point due to different forces acting together is the sum of the individual stresses due to individual forces alone.

Mathematically

Let the stress due to a force [tex]F_i[/tex] be [tex]\sigma _i[/tex]

and the stress due to combined forces be [tex]\sigma _f[/tex]

thus according to principle of superposition we have

[tex]\sigma_f=\sum_{i=1}^{n}(\sigma_i)[/tex]

The velocity gradients at y=0.25 ft and y=0.50 ft from the boundary are calculated by differentiating the velocity profile v = 4y^2/3 with respect to y and evaluating at the given points.

The velocity profile given by v = 4y2/3 describes the velocity of a fluid at different distances from the boundary in a flow field. To find the velocity gradient at the boundary (y=0), and at y=0.25 ft and y=0.50 ft, we differentiate the velocity profile with respect to y to get the gradient, dv/dy.

At the boundary, y=0, but since we're dealing with a power of y, the derivative will be a term that includes y in the denominator, which would imply an infinite gradient at the boundary, although physically this would manifest as a very large but finite value. However, at y=0.25 ft and 0.50 ft, we calculate the gradient by inserting these y-values into the derivative

Let's calculate the velocity gradient at y = 0.25 ft and y = 0.50 ft:

These calculations will give us the velocity gradients at the specified values of y, in ft/s2.

Designing the digital logic circuit for a security alarm involves using AND, OR, and NOT gates to allow any motion sensor to trigger the system, with a master switch for overall control and separate switches for the siren, lights, and automated call.

The question involves designing a digital logic circuit for a security alarm system with specific inputs and outputs. To achieve this configuration, one needs to employ AND, OR, and NOT gates to process signals from four motion sensors, a master switch, and individual enable switches for the siren, lights, and call function. The logic for this system requires that any motion sensor can trigger an alarm if the system is enabled, and each output (siren, lights, call) is individually controllable. Given the complexity of drawing a circuit diagram in text format, it's important to visualize the circuit as starting with OR gates to combine signals from the four motion sensors. These signals then pass through an AND gate along with the master switch to ensure the system is enabled. The outputs of this gate would then be directed to three separate AND gates, each also receiving input from their respective enable switch (for the siren, lights, and call functions). NOT gates may be used where necessary to invert signals, particularly for the enable/disable logic.

Answer:

1) Object C has the largest mass.

2)Object  has the smallest mass.

3) Weight of A = 26.6 Newtons

4)Weight of B = 8.6184 Newtons

5)Weight of C = 27.7286 Newtons

Explanation:

Since all the given masses have different unit's we shall convert them all into a same base unit for comparison. The base unit is selected to be kilogram.

Hence

1) Mass of object A = 7 kilogram

2) Mass of object B = 5 pounds

We know that 1 pound equals 0.4536 kilograms Hence 5 pounds equals

[tex]0.4536\times 5=2.268kg[/tex]

3) Mass of object C = 0.5 slug

We know that 1 slug equals 14.594 kilograms Hence 0.5 slug equals

[tex]14.594\times 0.5=7.297kg[/tex]

Upon comparing all the 3 masses we conclude that object C has the largest mass and object B has the smallest mass.

Part b)

Weight of an object is given by

[tex]Weight=mass\times g[/tex]

Now on Mars value of g equals [tex]3.8m/s^{2}[/tex]

Thus the corresponding weights are as under:

[tex]W_{A}=3.8\times 7=26.6N\\\\W_{B}=3.8\times 2.268=8.6184N\\\\W_{C}=3.8\times 7.297=27.7286N\\\\[/tex]