Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is 0.9 cm. The water density is 1.0 kg/L. The empty weight of the metal pipe is 2500 N/m. In kN, what is the total weight (pipe plus water)?

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 temperature attains equilibrium with the surroundings.  

Explanation:

When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to  constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.

Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.

Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.

Answer:

Pump power is 23.09 kW

Explanation:

Data

gravitational constant, [tex] g = 9.81 m/s^2 [/tex]

mass flow, [tex] \dot{m} = 60.6 kg/s [/tex]

flow density, [tex] \rho = 1000 kg/m^3 [/tex]

pump efficiency, [tex] \eta = 0.74 [/tex]

output gage pressure, [tex] p_o = 344.75 kPa [/tex]

input gage pressure, [tex] p_i = 68.95 kPa [/tex]

output pipe area, [tex] A_o = 0.069 m^2 [/tex]

input pipe area, [tex] A_i = 0.093 m^2 [/tex]

output height, [tex] z_o = 1.22 m - 0.61 m = 0.61 m [/tex] (considering that pump is at the maximum height, i.e., 1.22 m)

input height, [tex] z_i = 0 m [/tex]

pump hydraulic power,[tex] P = ? kW [/tex]

First of all, volumetric flow (Q) must be computed

[tex] Q = \frac{\dot{m}}{\rho}[/tex]

[tex] Q = \frac{60.6 kg/s}{1000 kg/m^3} [/tex]

[tex] Q = 0.0606 m^3/s[/tex]

Then, velocity (v) must be computed for both input and output

[tex] v_o = \frac{Q}{A_o}[/tex]

[tex] v_o = \frac{0.0606 m^3/s}{0.069 m^2}[/tex]

[tex] v_o = 0.88 m/s [/tex]

[tex] v_i = \frac{Q}{A_i}[/tex]

[tex] v_i = \frac{0.0606 m^3/s}{0.093 m^2}[/tex]

[tex] v_i = 0.65 m/s [/tex]

Now, total head (H) can be calculated

[tex] H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g} [/tex]

[tex] H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2} [/tex]

[tex] H = 28.74m [/tex]

Finally, pump power is computed as

[tex] P = \frac{Q \, \rho \, g \, H}{\eta}[/tex]

[tex] P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}[/tex]

[tex] P = 23.09 kW [/tex]

The hydraulic power in kW is mathematically given as

P = 23.09 kW

Generally the equation for the volumetric flow (Q)  is mathematically given as

Q=m/p

Therefore

Q=60/1000

Q=0.0606

Generally the equation for the velocity (v)    is mathematically given as

v=Q/A

Hence for input

[tex]v_i = \frac{Q}{A_i}\\\\v_i = \frac{0.0606 }{0.093 }[/tex]

v_i = 0.65 m/s

For input

[tex]v_o = \frac{Q}{A_o}\\\\v_o = \frac{0.0606 }{0.069}[/tex]

v_o = 0.88 m/s

Therefore, Total head (H)

[tex]H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\mu g}[/tex]

[tex]H = (0.61 0 ) + \frac{{0.88 }^2 - {0.65 }^}{2 *9.81 } + \frac{(344.75 Pa-68.95 Pa)*10^3}{1000* 9.81}[/tex]

H = 28.74m

Generally the equation for the pump power P  is mathematically given as

[tex]P = \frac{Q * \rho*g*H}{\eta}[/tex]

[tex]P = \frac{0.0606 * 100*9.81*28.74}{0.74}[/tex]

P = 23.09 kW

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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:

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:

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:

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]

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:

[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:

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:

1160 K.

Explanation:

Given that

Initial

Pressure P =600 KPa

Temperature T =290 K

Volume V =0.01 [tex]m^3[/tex]

If we assume that air is s ideal gas the

P V = mRT

R=0.287 KJ/kg.k

now by putting the values in above equation

600 x 0.01 = m x 0.287 x 290

m=0.07 kg

The work out at constant pressure given as

[tex]w=P(V_2-V_1)[/tex]

[tex]18=600(V_2-0.01)[/tex]

[tex]V_2=0.04\ m^3[/tex]

At constant pressure

[tex]\dfrac{T_2}{T_1}=\dfrac{V_2}{V_1}[/tex]

[tex]\dfrac{T_2}{290}=\dfrac{0.04}{0.01}[/tex]

[tex]T_2=1160\ K[/tex]

So the final temperature is 1160 K.

Answer:

The equations for magnitude and phase of current and voltages on resistor and inductor are:

[tex]I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]

[tex]V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]

[tex]V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}[/tex]

Explanation:

The first step is to find the impedances of the resistance (R) and the inductor (L).

The impedance of the resistor is:

The impedance of the inductor is:

Where [tex]\omega [/tex] is the angular frequency of the source, and the angle is [tex]90^{\circ} [/tex] because a pure imaginary number is on the imaginary axis (y-axis).

The next step is to find the current expression. It is the same for the resistor and inductor because they are in series. The total impedance equals the sum of each one.

[tex]I=\frac{V}{Z_R+Z_L}[/tex]

It is said that [tex]V=V_m\angle \theta[/tex], so, the current would be:

[tex]I=\frac{V_m\angle \theta }{R+j\omega L}[/tex]

The numerator must be converted to polar form by calculating the magnitude and the angle:

The current expression would be as follows:

[tex]I=\frac{V_m\angle \theta }{\sqrt{R^2+(\omega L)^2}\, \angle tan^{-1}(\omega L / R)}[/tex]

When dividing, the angles are subtracted from each other.

The final current expression is:

[tex]I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]

The last step is calculating the voltage on the resistor [tex]V_R[/tex] and the voltage on the inductor [tex]V_L[/tex]. In this step the polar form of the impedances could be used. Remember that [tex]V=I\cdot Z[/tex].

(Also remember that when multiplying, the angles are added from each other)

Voltage on the resistor [tex]V_R[/tex]

[tex]V_R=I\cdot Z_R=\bigg( \frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)\bigg) \cdot (R\angle 0^{\circ})[/tex]

The final resistor voltage expression is:

[tex]V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]

Voltage on the inductor [tex]V_L[/tex]

[tex]V_L=I\cdot Z_L=\bigg( \frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)\bigg) \cdot (\omega L \angle 90^{\circ})[/tex]

The final inductor voltage expression is:

[tex]V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}[/tex]

Summary: the final equations for magnitude and phase of current and voltages on resistor and inductor are:

[tex]I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]

[tex]V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]

[tex]V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}[/tex]

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.

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:

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]

Answer:

The correct answer is option 'b':0.25

Explanation:

By definition of strain we have

[tex]\epsilon =\frac{L_f-L_o}{L_o}[/tex]

where

[tex]\epsilon [/tex] is the strain

[tex]L_o[/tex] is the original length of specimen

[tex]L_f[/tex] is the elongated length of specimen

Applying the given values we get

[tex]\epsilon =\frac{12.5''-10''}{10''}=0.25[/tex]

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:

16.38L

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties.

Quality is defined as the ratio between the amount of steam and liquid when a fluid is in a state of saturation, this means that since the quality is 50%, 1kg is liquid and 1kg is steam.

then to solve this problem we find the specific volume for ammonia in a saturated liquid state at 20C, and multiply it by mass (1kg)

v(amonia at 20C)=0.001638m^3/kg

m=(0.01638)(1)=0.01638m^3=16.38L

Answer:

It is attempting to change the value of a constant.

Explanation:

In this pseudocode program "GRAVITY" is declared as a constant value, therefore it cannot be changed during runtime. If you tried mo write this code in a real language and compile it you would get a compilation error because of that forbidden operation.

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:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

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:

100Btu

Explanation:

According to the First law of thermodynamics:

ΔQ = W + ΔU

Where,  

ΔQ is the heat change

W is the amount of work done

ΔU  is change in internal energy

Since, there is no work done on the system as heat is just passing from coffee mug to surroundings, So, W = 0

Thus,

Net heat change = change in internal energy.

Change in internal energy = [tex]U_f-U_i=(68-168)\ Btu=-100\ Btu[/tex]

Thus,

ΔQ = -100 Btu  (negative sign indicates release of heat)

So,  

Heat flown to the room = 100Btu

Answer:

b)False

Explanation:

A battery is a device which store the energy in the form of chemical energy.And this stored energy is used according to the requirement.So battery is not a electromechanical device.Because it does have any mechanical component like gear ,shaft flywheel etc.

A flywheel is known as mechanical battery because it stored mechanical energy and supply that energy when more energy is required.Generally fly wheel is used during punching operation.

Answer:

[tex]\sigma = 65.25\ ksi[/tex]

Explanation:

given data:

D = 8 inch

R =4 inch

thickness = 0.018 inch

width = 0.625 inch

[tex]E =29*10^6 psi[/tex]

from bending equation we know that

[tex]\frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R}[/tex]

[tex]\sigma = \frac{Ey}{R}[/tex]

Where y represent distance from neutral axis

[tex]y = \frac{t}[2}[/tex]

[tex]y = \frac{0.018}{2}[/tex]

y = 0.009inch

[tex]\sigma = \frac{29*10^6*0.009}{4}[/tex]

[tex]\sigma = 65250\ psi[/tex]

[tex]\sigma = 65.25\ ksi[/tex]

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:

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:

a) temperature: random error

b) parallax: systematic error

c) using incorrect value: systematic error

Explanation:

Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.

Answer:

The dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.

Explanation:

Step1

Given:

Inner diameter is 2.00 in.

Gap between cups is 0.2 in.

Length of the cylinder is 2.5 in.

Rotation of cylinder is 10 rev/min.

Torque is 0.00011 in-lbf.

Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.

Step2

Calculation:

Tangential force is calculated as follows:

T= Fr

[tex]0.00011 = F\times(\frac{2}{2})[/tex]

F = 0.00011 lb.

Step3

Tangential velocity is calculated as follows:

[tex]V=\omega r[/tex]

[tex]V=(\frac{2\pi N}{60})r[/tex]

[tex]V=(\frac{2\pi \times10}{60})\times1[/tex]

V=1.0472 in/s.

Step4

Apply Newton’s law of viscosity for dynamic viscosity as follows:

[tex]F=\mu A\frac{V}{y}[/tex]

[tex]F=\mu (\pi dl)\frac{V}{y}[/tex]

[tex]0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}[/tex]

[tex]\mu =1.3374\times 10^{-6}[/tex]lb-s/in².

Step5  

Kinematic viscosity is calculated as follows:

[tex]\upsilon=\frac{\mu}{\rho}[/tex]

[tex]\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}[/tex]

[tex]\upsilon=1.4012\times 10^{-3}[/tex] in2/s.

Thus, the dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.