4. Which of these is typically NOT a function of the flywheel?
OA. Power transfer to the transmission
OB. Engage the Starter
OC. Smooth out power pulses
OD. Balance the crankshaft

Answer:

Quick Breads

Explanation:

I think this is correct. I am not for sure i just looked it up and it says word for word that this is the answer.

Answer:

Quick Breads

Explanation:

I got it right on my Plato test

Answer:

The relation between the discharge volume flow rate and the inlet volume flow rate is [tex]\frac{1}{5}[/tex].

Explanation:

No matter if fluid is compressible or not, mass throughout compressor, a device that works at steady state, must be conserved according to Principle of Mass Conservation:

[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (Eq. 1)

Where [tex]\dot m_{in}[/tex] and [tex]\dot m_{out}[/tex] are mass flows at inlet and outlet, measured in kilograms per second.

After applying Dimensional analysis, we expand the equation above as follows:

[tex]\rho_{in}\cdot \dot V_{in} - \rho_{out}\cdot \dot V_{out} = 0[/tex] (Eq. 2)

Where:

[tex]\rho_{in}[/tex], [tex]\rho_{out}[/tex] - Fluid densities at inlet and outlet, measured in kilograms per cubic meter.

[tex]\dot V_{in}[/tex], [tex]\dot V_{out}[/tex] - Volume flow rates at inlet and outlet, measured in cubic meters per second.

After some algebraic handling, we find the following relationship:

[tex]\rho_{out}\cdot \dot V_{out} = \rho_{in}\cdot \dot V_{in}[/tex]

[tex]\frac{\dot V_{out}}{V_{in}} = \frac{\rho_{in}}{\rho_{out}}[/tex] (Eq. 3)

If we know that [tex]\rho_{in} = 1\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{out} = 5\,\frac{kg}{m^{3}}[/tex], then the relation between the discharge volume flow rate and the inlet volume flow rate is:

[tex]\frac{\dot V_{out}}{\dot V_{in}} = \frac{1\,\frac{kg}{m^{2}} }{5\,\frac{kg}{m^{3}} }[/tex]

[tex]\frac{\dot V_{out}}{\dot V_{in}} = \frac{1}{5}[/tex]

The relation between the discharge volume flow rate and the inlet volume flow rate is [tex]\frac{1}{5}[/tex].

The phase identifying the need involves research to determine exactly what the client expects. The correct option is B.

Research is a systematic inquiry process that includes data gathering, documentation of important information, analysis, and interpretation of that data and information.

These all are in accordance with appropriate procedures established by particular academic and professional disciplines.

Action-informing research is its goal. As a result, your study should attempt to place its findings in the perspective of the wider body of knowledge. In order to develop knowledge that is usable outside of the research setting, research must constantly be of the highest calibre.

Research is done during the step of determining the need to ascertain the precise expectations of the client.

Thus, the correct option is B.

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Answer:

Usually it can be changed by flipping a switch. If your machine does not have one, then you just need to exchange the cables to the electrode holder and ground clamp.

Explanation:

The things that must be done before you change the polarity on a welding machine are coarse and fine adjustment.

The heated tool assembly, with two exposed surfaces, and two fixtures for holding the component parts to be welded. The tooling is for bringing the component parts into contact with the heated tool and bringing the molten joint surfaces together to form the weld. And displacement stops on the platen and holding fixtures make up a heated tool welding machine.

Polarity refers to the fact that the electrical circuit formed when you turn on the welder contains a negative and a positive pole. When welding, polarity is crucial, since the weld's strength and quality are affected by the choice of polarity.

Therefore, before changing the polarity of welding equipment, coarse and fine adjustments must be made.

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Answer:

The wind turbine generates [tex]19297.222[/tex] kilowatt-hours of electricity daily.

The wind turbine makes a daily revenue of 1736.75 US dollars.

Explanation:

First, we have to determine the stored energy of wind ([tex]E_{wind}[/tex]), measured in Joules, by means of definition of Kinetic Energy:

[tex]E_{wind} = \frac{1}{2}\cdot \dot m_{wind}\cdot \Delta t \cdot v_{wind}^{2}[/tex] (Eq. 1)

Where:

[tex]\dot m_{wind}[/tex] - Mass flow of wind, measured in kilograms per second.

[tex]\Delta t[/tex] - Time in which wind acts in a day, measured in seconds.

[tex]v_{wind}[/tex] - Steady wind speed, measured in meters per second.

By assuming constant mass flow and volume flows and using definitions of mass and volume flows, we expand the expression above:

[tex]E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot \dot V_{air} \cdot \Delta t \cdot v_{wind}^{2}[/tex] (Eq. 1b)

Where:

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\dot V_{air}[/tex] - Volume flow of air through wind turbine, measured in cubic meters per second.

[tex]E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot A_{c}\cdot \Delta t\cdot v_{wind}^{3}[/tex] (Eq. 2)

Where [tex]A_{c}[/tex] is the area of the wind flow crossing the turbine, measured in square meters. This area is determined by the following equation:

[tex]A_{c} = \frac{\pi}{4}\cdot D^{2}[/tex] (Eq. 3)

Where [tex]D[/tex] is the diameter of the wind turbine blade, measured in meters.

If we know that [tex]\rho_{air} = 1.25\,\frac{kg}{m^{3}}[/tex], [tex]D = 100\,m[/tex], [tex]\Delta t = 86400\,s[/tex] and [tex]v_{wind} = 8\,\frac{m}{s}[/tex], the stored energy of the wind in a day is:

[tex]A_{c} = \frac{\pi}{4}\cdot (100\,m)^{2}[/tex]

[tex]A_{c} \approx 7853.982\,m^{2}[/tex]

[tex]E_{wind} = \frac{1}{2}\cdot \left(1.25\,\frac{kg}{m^{3}} \right) \cdot (7853.982\,m^{2})\cdot (86400\,s)\cdot \left(8\,\frac{m}{s} \right)^{3}[/tex]

[tex]E_{wind} = 2.171\times 10^{11}\,J[/tex]

Now, we proceed to determine the quantity of energy from wind being used by the wind turbine in a day ([tex]E_{turbine}[/tex]), measured in joules, with the help of the definition of efficiency:

[tex]E_{turbine} = \eta\cdot E_{wind}[/tex] (Eq. 4)

Where [tex]\eta[/tex] is the overall efficiency of the wind turbine, dimensionless.

If we get that [tex]E_{wind} = 2.171\times 10^{11}\,J[/tex] and [tex]\eta = 0.32[/tex], then the energy is:

[tex]E_{turbine} = 0.32\cdot (2.171\times 10^{11}\,J)[/tex]

[tex]E_{turbine} = 6.947\times 10^{10}\,J[/tex]

The wind turbine generates [tex]6.947\times 10^{10}[/tex] joules of electricity daily.

A kilowatt-hours equals 3.6 million joules. We calculate the equivalent amount of energy generated by wind turbine in kilowatt-hours:

[tex]E_{turbine} = 6.947\times 10^{10}\,J\times\frac{1\,kWh}{3.6\times 10^{6}\,J}[/tex]

[tex]E_{turbine} = 19297.222\,kWh[/tex]

The wind turbine generates [tex]19297.222[/tex] kilowatt-hours of electricity daily.

Lastly, the revenue generated per day can be found by employing the following:

[tex]C_{rev} = c\cdot E_{turbine}[/tex] (Eq. 5)

Where:

[tex]c[/tex] - Unit price, measured in US dollars per kilowatt-hour.

[tex]C_{rev}[/tex] - Revenue generated by the wind turbine in a day, measured in US dollars.

If we know that [tex]c = 0.09\,\frac{USD}{kWh}[/tex] and [tex]E_{turbine} = 19297.222\,kWh[/tex], then the revenue is:

[tex]C_{rev} = \left(0.09\,\frac{USD}{kWh} \right)\cdot (19297.222\,kWh)[/tex]

[tex]C_{rev} = 1736.75\,USD[/tex]

The wind turbine makes a daily revenue of 1736.75 US dollars.

Answer:

Colloids are very low diameter particles which are responsible for the turbidity or the color of surface water. Because of their very low sedimentation speed the best way to eliminate them is the coagulation-flocculation processes.

Answer: the total chronic daily intake is  0.03296 mg/ kg.d  

Explanation:

First of  we have to determine the CDI due to ingestion of drinking water

so

CDI = (C x IR x EF x ED) / (BW x AT)     let leave this as equation 1

where C is the concentration of the life time risk of drinking water(1.0mg/L)

ED is the risk dying per year((70),

EF is the number of days per year(365 day/year)

BW is the weight of body(65.4)kg

AT is the average tenure time of life(75years)

IR = 2.3 L/days

so we substitute

CDI₁ = (1(mg/L) x 2.3(L/day) x 365 x 70) / (65.4 x 365 x 75)

= 3.29 x 10⁻² mg/kg.d  

next we determine the CDI due to dermal contact with water.  

find the ingestion rate which is equal to the exposure time in hour/day.

IR = ET

(20 min/day) /  (60 min/hour)

= 0.333 h/day

now lets substitute for CD1₂

1.0 mg/L for C , 0.333 h/day for m , 365 day/year for EF, 70 year for ED, 65.4 kg for Bw , and 75 year for AT

0.8 for submergence and 1.69m² for area of skin of adult female in our equation 1

CDI₂ = [(1 mg/1) × (1.69 m²) × (9.0 x 10⁻⁶(m/h)) × 0.333(h/day) x 365 x 70)) / (65.4 x 365 x 75)]  × ( 0.8 x 10³ L/m³)

CDI₂ = 5.41 x 10⁻⁵ mg/kg.d  

now we find the CDI due to inhalation during bath

we substitute 1.0 mg/L for C, 11.3m³/day for IR , 365 day/year for EF , 70 year for ED. 65.4 kg for BW , and 75 year for AT in our equation 1

CDI₃ = [( 1ug/m³ × 1mg/10³ug) x (11.3m³/day x 1day/24 hr) x 365 x 70)] / (65.4 kg x 365 x 75 )

= 6.71 x 10⁻⁶ mg/kg.d

finally we Calculate the total chronic daily intake value

CDI_total = CDI₁ + CDI₂ + CDI₃  

we substitute

CDI_total = 3.29 x 10⁻² + 5.41 x 10⁻⁵ + 6.71 x 10⁻⁶

= 0.03296 mg/kg.d  

so the total chronic daily intake is  0.03296 mg/ kg.d  

Answer:

a

Explanation:

Answer:

Explanation:

From the information given:

a.

Using the equipartition theorem, the average energy of a molecule dor each degree of freedom is:

[tex]U = \dfrac{1}{2}k_BT[/tex]

[tex]U = \dfrac{1}{2}nRT[/tex]

For s degree of freedom

[tex]U = \dfrac{1}{2}snRT[/tex]

However, the molar specific heat [tex]C_v = \dfrac{1}{n} \dfrac{dU}{dT}[/tex]

Therefore, in terms of R and s;

[tex]C_v = \dfrac{1}{n} \dfrac{d}{dT} \begin{pmatrix} \dfrac{1}{2} snRT \end {pmatrix}[/tex]

[tex]C_v = \dfrac{Rs}{2}[/tex]

b.

Given that:

Cv=70.6Jmol⋅K and R=8.314Jmol⋅K

Then; using the formula  [tex]C_v = \dfrac{Rs}{2}[/tex]

[tex]70.6 \ J/mol.K = \dfrac{(8.314 \ J/mol.K)\times s}{2}[/tex]

[tex]70.6 \ J/mol.K \times 2= (8.314 \ J/mol.K)\times s[/tex]

[tex]s= \dfrac{70.6 \ J/mol.K \times 2}{ (8.314 \ J/mol.K) }[/tex]

s = 16.983

s [tex]\simeq[/tex] 17

Answer:

Follows are the code to this question:

n=input("Enter the first 12 digits of an ISBN-13 as a string:")#defining a varaible isbn for input value

if len(n)!=12: #use if block to check input value is equal to 12 digits

   print("incorrect input") #print error message

elif n.isdigit()==False: #use else if that check input is equal to digit

   print("incorrect input") #print error message

else:# defining else block

   s=0 #defining integer vaiable s to 0

   for i in range(12):#defining for loop to calculate sum of digit

       if i%2==0: #defining if block to check even value

           s=s+int(n[i])#add even numbers in s vaiable  

       else: #use else block for odd numbers

           s=s+int(n[i])*3 #multiply the digit with 3 and add into s vaiable

   s=s%10#calculate the remainder value  

   s=10-s#subtract the remainder value with 10 and hold its value

   if s==10: #use if to check s variable value equal to 10  

       s=0#use s variable to assign the value 0

   n=n+s.__str__() #u

Output:

please find attached file.

Explanation:In the above Python code, the "n" variable is used for input the number into the string format uses multiple conditional statements for a check input value, which can be defined as follows:

Answer: maximum length of the nanowire is 510 nm

Explanation:

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m  

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 )  = ( (4h/kD)^1/2 )  

m =  ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04    

now lets find the value of h/mk    

h/mk = 10⁵ / ( 942809.04 × 30) =  0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb =  (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ 1 ]  / [cosh mL+  (h/mk) sinh mL]

so we substitute

0.893 =  [ 1 ]  / [cosh (942809.04 × L) +  (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

Answer:

Mechanic: a person who repairs and maintains machinery

Mechanical engineers: design power-producing machines

Explanation:

Answer: the lifetime (75 years) CDI is 0.001393 mg/kg.day

Explanation:

Given that;

CA = Contaminant Concentration = 0.05 mg/m³

IR = Inhalation Rate = 15.2m³per day = 15.2 / 24 = 0.6333 m³/hr

ET = Exposure Time = 8 hour per day

ED = Exposure Duration = (65 - 18 years)  = 47

EF = Exposure Frequency = 5 day/week * 50 week/ year = 250 days/year

BW = Body Weight = 78 Kg

AT = Averaging Time = 75 years = (75 * 365 days) = 27375 days

Now to find the CDI (Chronic Daily Intake)' we say;

CDI = (CA×IR×ET×EF×ED) / ( BW×AT)

so we substitute;

CDI = (0.05 × 0.6333 × 8 ×250 × 47) / (78 × 27375)

CDI = 2976.51 / 2135250

CDI = 0.001393 mg/kg.day

therefore the lifetime (75 years) CDI is 0.001393 mg/kg.day

Answer: Could you put this in english plz

Explanation:

Answer:

Fluid dynamics.

Materials science.

Robotics.

Manufacturing processes.

Thermodynamics and heat transfer.

Environmental science.

Explanation:

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

Answer: 11470.4 cubic yards

Explanation:

first we calculate the volume of the site;

V1 = Area × depth

V1 = 2.5 acre ×  (43560 ft² / 1 acre )×3ft

V1 = 326700 ft

next we is the relative compaction

RC = [γd(field) / γdmax(laboratory)] × 100

so we substitute

93 =  [γd(field) / 122 lb/ft³)] × 100

γd(field) = 113.46 lb/ft³

then the dry unit weight of the site

γday1 = γavg / ( 1 + w)

= 118 lb/ft³ / ( 1 + (9.6/100))

= 118 lb/ft³ /1.096

= 107.664 lb/ft³

finally we find the fill volume of the site

V2/V1 = γd / γd(field)

we substitute

V2/326700 = 107.664 / 113.46

V2 = 310010.83 ft³

we convert to cubic yards

= 310010.83 ft³ × (0.037 cubic yard / 1 ft³)

= 11470.4 cubic yards

Answer:

[a] False.

[b]. True

[c]. false.

[d]. true.

[e]. true.

Explanation:

NB: SDOF simply means Simple Degree Of Freedom, that is to say it is a system that can be solved by differential equation such as the second order and the single differential equation.

So, the question asked us to determine if each of the scenario is true or false.  

a. Consider a SDOF system with Coulomb damping. When displaced from the equilibrium position and released, the mass may not move at all.

ANSWER:  FALSE.

REASON: The mass moved a little bit When displaced from the equilibrium position and released.

b. Consider a SDOF system with an ideal viscous damper. When displaced from the equilibrium position and released, the mass will always undergo oscillatory motion.

ANSWER: TRUE

REASON:  for an ideal viscous damper. When displaced from the equilibrium position and released, the mass will always undergo oscillatory motion.

c. The damped natural frequency of a system is always greater than the undamped natural frequency

.

ANSWER:  FALSE

REASON:  The damped natural frequency of a system is always greater than the undamped natural frequency

d. For an undamped system undergoing a harmonic forcing, the amplitude of the response approaches zero as the forcing frequency becomes very high.

ANSWER: TRUE

REASON:  the amplitude of the response approaches zero as the forcing frequency becomes very high for  undamped system undergoing a harmonic forcing

e. The amplitude of free response of SDOF system with Coulomb damping decreases

ANSWER: TRUE

REASON:  amplitude of free response of SDOF system with Coulomb damping decreases

Answer:

Ribosomes 1

Explanation

Ribosomes are small structures where proteins are made. Although they are not enclosed within a membrane, they are frequently considered organelles. Each ribosome is formed of two subunits, like the one pictured at the top of this section. Both subunits consist of proteins and RNA

Answer:

The answer is "[tex]\bold{ 259.2 \times 10^{11} }[/tex]".

Explanation:

The amount of kilograms, which travel in a thick sheet of hydrogen:

[tex]M= -DAt \frac{\Delta C}{ \Delta x} \\\\[/tex]

[tex]D =1.0 \times 10^{8} \ \ \ \frac{m^2}{s} \\\\ A = 0.20 \ m^2\\\\t = 1\ \ h = 3600 \ \ sec \\\\[/tex]

calculating the value of [tex]\Delta C:[/tex]

[tex]\Delta C =C_A -C_B[/tex]

  [tex]= 2.4 - 0.6 \\\\ = 1.8 \ \ \frac{kg}{m^3}[/tex]

calculating the value of [tex]\Delta X:[/tex]

[tex]\Delta x = x_{A} -x_{B}[/tex]

     [tex]= 0 - (5\ mm) \\\\ = - 5 \ \ mm\\\\= - 5 \times 10^{-3} \ m[/tex]

[tex]M = -(1.0 \times 10^{8} \times 0.20 \times 3600 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\[/tex]

    [tex]= -(1.0 \times 10^{8} \times 720 \times (\frac{1.8}{-5 \times 10^{-3}})) \\\\= -(1.0 \times 10^{8} \times \frac{ 1296}{-5 \times 10^{-3}})) \\\\= (1.0 \times 10^{8} \times 259.2 \times 10^3)) \\\\= 259.2 \times 10^{11} \\\\[/tex]

Answer:

The answer "K = 0.0075"

Explanation:

If we try to measure up to 69 kPa of air, find mercury or fluid for gauge.  

While mercury was its largest liquid with a density of 13600 kg / m3 at normal room temperature.  

Let's all measure for 69 kPa that height of the  mercury liquid column.

[tex]\to P = 69 \ kPa[/tex]

       [tex]= 69000 Pa \\\\[/tex]

[tex]\to \rho = 13600 \ \ \frac{kg}{m^3} \\\\\\to g = 9.81 \ \ \frac{m}{s^2} \\\\[/tex]

Formula:

[tex]\to P=\rho \ gh[/tex]

[tex]\to 69000 = 13600\times9.81 \times h\\\\\to h= \frac{69000}{13600\times9.81} \\\\\to h= \frac{69000}{133416} \\\\\to h= 0.517179349 \\\\ \to h= 517 \ mm \\\\[/tex]

The right choice for pressure measurements up to 69 kPa is mercury.  

Atmospheric Mercury up to 69 kPa Air 517 mm  

The relationship of Hg to Pa is = 134.22 Pa 1 mm Hg  

Static sensitivity to Pa of mm hg = change of mercury height to Pa:

[tex]= \frac{\Delta Hg }{ \Delta P }\\\\= \frac{1 }{ 133.3 }\\\\= 0.0075[/tex]

Answer:

C: Viscosity, the resistance to flow that fluids exhibit

Explanation:

Did it on Edge :)

Answer:

C: Viscosity, the resistance to flow that fluids exhibit

Explanation: