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.
What is a welding machine?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.
To learn more about welding machines, refer to the link:
https://brainly.com/question/14272269
#SPJ2
All of the following are common causes of low engine compression except for
A. too much oil
B. defective valves.
C. weak piston rings.
D. bent pushrods
Answer:
B
Explanation:
Over filling of oil on an engine will not cause loss of compression. The crankshaft splashes it inside the engine and will come out the dip stick tube and breather on valve cover and it may splash up to the bottom of the pistons and force oil to the top through the rings
One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher-temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15-nm-diameter silicon carbide nanowire onto a silicon carbide sururface. The surface is maintained at a temperature of Ts = 2400 K and the particular liquid catalyst that is used must be maintained in the range 2400 K ≤ Tc ≤ 3000 K to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by h = 105 W/m2.K and T[infinity] = 8000 KT. Assume properties of the nanowire are the same as for bulk silicon carbide.
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
The site earthwork in Phase III of Four Hills Landfill project included a 3.0 ft deep cut across an entire 2.5-acre site. Soil was excavated from within the proposed Phase III footprint. The average unit weight of this soil is 118 lb/ft3, and the average moisture content is 9.6%. It also has a maximum dry unit weight of 122 lb/ft3 and an optimum moisture content of 11.1%, based on the modified Proctor test. The excavated soil will be placed on a nearby site and compacted to an average relative compaction of 93%. Compute the volume of fill that will be produced and express your answer in cubic yards.
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
...........................................................................................................
The weight of your car will aslo affect its
Answer:
speed and acceleration
A work cell is currently operated 2000 hr/yr by a human worker who is paid an hourly rate of $23.00, which includes applicable overhead costs. One work unit is produced in a cycle time of 4.8 min. Management would like to increase output to meet increasing demand and a robot cell is being considered as a replacement for the present manual cell. The cycle time of the proposed cell would be reduced to 4.0 min. The installed cost of the robot plus supporting equipment is $120,000. Power and other utilities to operate the robot will be $0.30/hr, and annual maintenance costs are $2500. Determine:
a. the number of parts produced annually by the manually operated cell.
b. cost per part produced.
c. how does the cost per part of the robot cell compare with your answer in part (b), given a 4-year service life, 10% rate of return, and no salvage value.
Answer:
a) 25000 pcs/yr
b) cost per part produced is $ 1.84 per pc
c) Cpc = $ 1.365 per pcs
Explanation:
a)
the relation to calculate the number of parts produced annually by manual process is;
Q = Hw / Tc
Hw is the hourly rate ( 2000hr/yr) and Tc is the cycle time ( 4.8 min)
so we substitute
Q = (2000 × 60) / 4.8
= 120000 / 4.8
= 25000 pcs/yr
b)
cost per part produced
the relation to calculate the cost per part produced is expressed as;
Cpc = $23(Hw) / Q
Cpc is the cost per part produced
so we substitute
Cpc = $23(2000) / 25000
= 46000 / 25000
= $ 1.8 per pc
therefore cost per part produced is $ 1.84 per pc
c)
for the robot cell, at a service life of 4 years and a 10% rate of return , the factor is expressed as;
f = [r(1 + r)^t] / [((1 + r)^t ) - 1 ]
our rate r = 10% = 0.1 and our t = 4
so we substitute
f = [0.1 (1 + 0.1)^4] / [((1 + 0.1)^4 ) - 1 ]
f = 0.14641 / 0.4641
f = 0.3155
now we find the total cost
TC = 120000(0.3155) + 2000(0.3) + 2500
TC = $ 40,960
next we find the parts produced annually
Q = (2000 × 60) / 4
Q = 120,000 / 4
Q = 30000 pcs/yr
finally we find the cost per part produced;
Cpc = TC / Q
we substitute
Cpc = 40,960 / 30000
Cpc = $ 1.365 per pcs
Pls help I will mark brainlesst
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
Complete the sentence using the correct term.
are made without yeast and are often savory or sweet.
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
A compressible fluid flows through a compressor that increases the density from 1 kg/m3 to 5 kg/m3. The cross-sectional area of the inlet pipe is 3 m2 and that of the discharge pipe is 1 m2. The relation between the discharge volume flow rate and the inlet volume flow rate is
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].
could you put your sparkplug in the steering wheel and it still works
A. no
B. yes
C. do not care
D. it really doesn't matter
Answer:
a
Explanation:
Assignment 1: Structural Design of Rectangular Reinforced Concrete Beams for Bending
Perform structural design of a rectangular reinforced concrete beam for bending. The beam is simply supported and has a span L=20 feet. In addition to its own weight the beam should support a superimposed dead load of 0.50 k/ft and a live load of 0.65 k/ft. Use a beam width of 12 inches. The depth of the beam should satisfy the ACI stipulations for minimum depth and be proportioned for economy. Concrete compressive strength f’c = 4,000 psi and yield stress of reinforcing bars fy = 60,000 psi. Size of stirrups should be chosen based on the size of the reinforcing bars. The beam is neither exposed to weather nor in contact with the ground, meaning it is subjected to interior exposure.
• Use the reference on "Practical Considerations for Rectangular Reinforced Concrete Beams"
• Include references to ACI code – see slides from second class
• Include references to Tables from Appendix A
• Draw a sketch of the reinforced concrete beam showing all dimensions, number and size of rebars, including stirrups.
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.
which one of these reduce fraction?
Mechanic... Mechanical Engineer... What's the difference?
Instructions: Answer the question below with at least TWO complete sentences.
Answer:
Mechanic: a person who repairs and maintains machinery
Mechanical engineers: design power-producing machines
Explanation:
Select all the correct answers. What are two reasons why the terrestrial planets formed closer to the sun after a supernova event that initiated the formation of the solar system?
Create an array of 10 size and assign 10 random numbers. Now find the sum of the array using for and while loop.
Answer:
10
Explanation:
Given int variables k and total that have already been declared, use a do...while loop to compute the sum of the squares of the first 50 counting numbers, and store this value in total. Thus your code should put 1*1 + 2*2 + 3*3 +... + 49*49 + 50*50 into total. Use no variables other than k and total.
Answer:
k-1000
Explanation:
Answer:
k = 1;
total = 0;
do {
total = total + (k*k);
k++;
} while (k <= 50);
Explanation:
answer using Java
Chad is working on a design that uses the pressure of steam to control a valve in order to increase water pressure in showers. Which statement describes what his design is dealing with? viscosity, the resistance to flow that fluids exhibit elasticity, the indicator of how easily a fluid will return to its original volume after a force is removed from the fluid compressibility, the property that describes the amount of volume that decreases when pressure is applied to a fluid hydraulics, the use and study of liquid fluids and their motions, behaviors, and interactions
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:
La distancia que existe entre las bases de un campo de béisbol es de 28 m. Si la pelota se batea por la línea en dirección a la tercera base con una velocidad de 32 m/s. ¿Con qué rapidez cambia la distancia entre la pelota y la primera base cuando se encuentra a la mitad del camino hacia la tercera base?
Answer: Could you put this in english plz
Explanation:
Design a U-tube manometer that can measure gage pressures up to 69 kPa of air. You will want to choose a manometer fluid with good static sensitivity but will not result in an unreasonably tall manometer. Further, the manometer fluid should be mostly immiscible with the air. The two design parameters you should consider are manometer fluid (impacts manometer fluid density) as well as the manometer height.
Required:
Compute the static sensitivity, K, in mmHg/Pa
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]
Estimate the chronic daily intake of toluene from exposure to a city water supply that contains a toluene concentration equal to the drinking water standard of 1 mg · L−1. Assume the exposed individual is an adult female who consumes water at the adult rate for 70 years, that she abhors swimming, and that she takes a long (20 minute) bath every day. Assume that the average air concentration of toluene during the bath is 1 µg · m−3. Assume the dermal uptake from water (PC) is 9.0 × 10−6 m · h−1 and that direct dermal absorption during bathing is no more than 80% of the available toluene because she is not completely submerged. Use the EPA lifetime exposure of 75 years.
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
Which courses might a mechanical engineer take in college
Answer:
Fluid dynamics.
Materials science.
Robotics.
Manufacturing processes.
Thermodynamics and heat transfer.
Environmental science.
Explanation:
Large wind turbines with blade span diameters of over 100 m are available for electric power generation. Consider a wind turbine with a blade span diameter of 100 m installed at a site subjected to steady winds at 8 m/s. Taking the overall efficiency of the wind turbine to be 32 percent and the air density to be 1.25 kg/m3 , determine the electric power generated by this wind turbine. Also, assuming steady winds of 8 m/s during a 24-hour period, determine the amount of electric energy and the revenue generated per day for a unit price of $0.09/kWh for electricity
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.
Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m^2 at 500°C. Assume a diffusion coefficient of 1.0 x 10^8 m^2 /s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.
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]
Thermodynamics deals with the macroscopic properties of materials. Scientists can make quantitative predictions about these macroscopic properties by thinking on a microscopic scale. Kinetic theory and statistical mechanics provide a way to relate molecular models to thermodynamics. Predicting the heat capacities of gases at a constant volume from the number of degrees of freedom of a gas molecule is one example of the predictive power of molecular models. The molar specific heat Cv of a gas at a constant volume is the quantity of energy required to raise the temperature T of one mole of gas by one degree while the volume remains the same. Mathematically, Cv=1nΔEthΔT, where n is the number of moles of gas, ΔEth is the change in internal (or thermal) energy, and ΔT is the change in temperature. Kinetic theory tells us that the temperature of a gas is directly proportional to the total kinetic energy of the molecules in the gas. The equipartition theorem says that each degree of freedom of a molecule has an average energy equal to 12kBT, where kB is Boltzmann's constant 1.38×10^−23J/K. When summed over the entire gas, this gives 12nRT, where R=8.314Jmol⋅K is the ideal gas constant, for each molecular degree of freedom.
Required:
a. Using the equipartition theorem, determine the molar specific heat, Cv , of a gas in which each molecule has s degrees of freedom. Express your answer in terms of R and s.
b. Given the molar specific heat Cv of a gas at constant volume, you can determine the number of degrees of freedom s that are energetically accessible. For example, at room temperature cis-2-butene, C4H8 , has molar specific heat Cv=70.6Jmol⋅K . How many degrees of freedom of cis-2-butene are energetically accessible?
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
NEEDS TO BE IN PYTHON:ISBN-13 is a new standard for indentifying books. It uses 13 digits d1d2d3d4d5d6d7d8d910d11d12d13 . The last digit d13 is a checksum, which is calculated from the other digits using the following formula:10 - (d1 + 3*d2 + d3 + 3*d4 + d5 + 3*d6 + d7 + 3*d8 + d9 + 3*d10 + d11 + 3*d12) % 10If the checksum is 10, replace it with 0. Your program should read the input as a string. Display "incorrect input" if the input is incorrect.Sample Run 1Enter the first 12 digits of an ISBN-13 as a string: 978013213080The ISBN-13 number is 9780132130806Sample Run 2Enter the first 12 digits of an ISBN-13 as a string: 978013213079The ISBN-13 number is 9780132130790
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:
In if block, it checks the length isn't equal to 12, if the condition true, it will print an error message. In the else, if the block it checks input value does not digit, if the condition is true, it will print an error message. In the else block, it uses the for loop, in which it calculates the even and odd number sum, and in the odd number, we multiply by 3 then add into s variable. In this, the s variable is used to calculate its remainder and subtract from the value and use the if block to check, its value is not equal to 10 if it's true, it adds 0 into the last of n variable, otherwise, it adds its calculated value.The recommended time weighted average air concentration for occupational exposure to water soluble hexavalent chromium (Cr VI) is 0.05 mg · m−3. This concentration is based on an assumption that the individual is generally healthy and is exposed for 8 hours per day, 5 days per week, 50 weeks per year, over a working lifetime (that is from age 18 to 65 years). Assuming a body weight of 78 kg and inhalation rate of 15.2 m3 · d−1 over the working life of the individual, what is the lifetime (75 years) CDI?
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
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:D
Explanation:
Option D is correct. Balancing the crankshaft is NOT one of the functions of a flywheel.
A flywheel is part of an automobile. It is a heavy wheel that is attached to a rotating shaft of an automobile. Since there may be the presence of large forces within an automobile structure, the presence of a flywheel helps to balance the force evenly across it.
Some of the functions of the flywheel are but are not limited to help in transferring power to the transmission, engaging the starter, and smoothing out of power pulses.
Hence based on the explanation above, balancing the crankshaft is NOT one of the function of a flywheel.
Learn more on flywheel here: https://brainly.com/question/14273307
what is the impact of colloidal particles in water
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.
Which phase involves research to determine exactly what the client expects?
brainstorming
identifying the need
preventive maintenance
building a model
The phase identifying the need involves research to determine exactly what the client expects. The correct option is B.
What is research?
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.
For more details regarding research, visit:
https://brainly.com/question/18723483
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State whether the following statements are true or false. Provide a brief justification for your answer.
a. Consider a SDOF system with Coulomb damping. When displaced from the equilibrium position and released, the mass may not move at all.
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.
c. 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.
e. The amplitude of free response of SDOF system with Coulomb damping decreases
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