Answer: the measurement of pressure of the order of 100 KPa to an accuracy of +- 5Kpa is POSSIBLE with the sensor
Explanation:
Given that;
specifications include the range 70 kPa to 1 MPa which includes 100 kPa.
specification include non-linearity and hysteresis error of 0.5%
first we find the non-linear error
= 1 MPa * 0.5%
= 1 MPa * 0.005
= 0.005 MPa = 5 KPa
therefore, it includes the non-linearity error of +- 5 KPa.
the specification include the temperature range -54 to 120 degree Celsius which includes 20 degree Celsius
THEREFORE, the measurement of pressure of the order of 100 KPa to an accuracy of +- 5Kpa is POSSIBLE with the sensor
what is a possible negative aspect of economic growth
Explanation:
Increased pollution is a possible negative aspect of economic growth
Kris and James are working at a construction site that has a significant amount of stagnant water. Which type of hazard are they most likely to be exposed to?
Answer:
A biological hazard
Explanation:
Biological because insects and other organisms thrive in stagnant water.
The type of hazard that is most likely to be exposed to a significant amount of stagnant water is known as biological.
What is meant by biological hazard?A biological hazard may be defined as a biological substance that may significantly pose a great threat to the health of living organisms, primarily humans. These types are the major concerns in food processing because they cause most foodborne illness outbreaks.
In the case of Kris and James, they are significantly exposed to a biological hazard because stagnant water is commonly utilized by mosquitos to place eggs, this directs a lot of mosquitos around stagnant waters and therefore a higher risk of mosquito-transmitted diseases such as malaria. Besides this, stagnant water is highly polluted and includes bacteria and parasites that are harmful.
Therefore, the type of hazard that is most likely to be exposed to a significant amount of stagnant water is known as biological.
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Large quantities of liquefied natural gas (LNG) are shipped by ocean tanker. At the unloading port, provision is made for vaporization of the LNG so that it may be delivered to pipelines as gas. The LNG arrives in the tanker at atmospheric pressure and 113.7 K and represents a possible heat sink for use as the cold reservoir of a heat engine. Assuming unloading of LNG as a vapor at the rate of 8,000 m3s-1, as measured at 25 C and 1.0133 bar, and assuming the availability of an adequate heat source at 35 C, what is the maximum amount of work that could be generated and what is the rate of heat transfer from the heat source
Answer:
yfyuhvckydfxcvkjyfvgrjfvhkhgfyhtfhchghjgyfvyghvhygvkuh
Explanation:
what are three common types of variable resistors
Answer:
There are 3 main types of resistors based on their composition: carbon-composition resistors, carbon-film resistors, and metal-film resistors.
Explanation:
there is also
Potentiometer.
Rheostat.
Thermistor.
Magneto resistor.
Photoresistor.
Humistor.
Force sensitive resistor.
Which of the following scenarios describes someone who is a materials engineer?
Jon is helping to create a new blood pressure arm cuff.
Kristin is currently modifying a newly discovered plastic material to work on a new lightweight lacrosse stick.
Thomas uses a computer program to be sure that all of the architect’s blueprints are structurally sturdy.
Ethan is helping to design a new bug-resistant soil for garden beds.
Answer: Kristin is currently modifying a newly discovered plastic material to work on a new lightweight lacrosse stick.
Explanation:
The main function of a materials engineer is to develop, study and test materials that are used on order to make different products.
Material engineer solve problems in other engineering fields, like electrical, aerospace, civil, mechanical, chemical, and nuclear.
From the information given in the question, the correct option is "Kristin is currently modifying a newly discovered plastic material to work on a new lightweight lacrosse stick."
Answer:
nice
Explanation:
might give brainliest
Complete the following sentence.
_______ is a communication system that is composed of images that represent words.
Answer:
Blissymbols
Explanation:
Blissymbols is a constructed language with hundreds of basic symbols that represent words.
Answer:
Blissymbols
Explanation:
Lake Superior, the largest of the Great Lakes of North America, is also the world's largest freshwater lake by surface area (~ 82,100 km2), and the third largest freshwater lake by volume (~ 12,100 km3). The monthly average precipitation in 2018 in the lake area was 2.69 inches. According to the data provided by USGS water stations, the total flow rate in all the incoming streams was found to be 356,747 ft3/s (cubic feet per second). At the same time, the lake discharges 717,258 ft3/s to its surrounding water bodies. The monthly average evaporation was 18.7 mm. The groundwater inflow was 783.33 km3 less than the groundwater outflow in this year. (Hints: (1) The average monthly data can be used to calculate the annual data; (2) Volume = Area * Height; (3) Pay special attention to the units, convert all the units to be consistent first.)
(1) With the information above, please write the water budget for Lake Superior (10 points);
(2) Please estimate the change of storage (in m3) in the year of 2018 (20 points);
(3) Please estimate the increase or decrease of water level (in mm) in the year of 2018 (10 points).
Answer:
1) V_win = 7.05 10¹⁰ m³ , 2)V_ lost = 1.8471 10¹⁰ m³, 3) h = 633.4 mm
Explanation:
In this exercise we will start by reducing all units to the SI system
h1 = 2.69 inch (2.54 10-2 m / 1 inch) = 6.8326 10⁻² m
Ф1 = 356.747 ft3 / s (1 m3 / 35.3147 ft3) = 10.1019 m³ / s
Ф2 = 717,258 ft3 / s = 20,310 m³ / s
h2 = 18.7 mm (1 m / 1000 mm) = 18.7 10⁻³ m
V_subterránea = - 783.33 10³ m,
Now let's answer the questions
1) they ask us the amount of water that reaches the lake in 2018
volume of rainwater is
V₁ = A. h t
V₁ = 82 100 106 * 6.8326 10-2 * 12
V₁ = 6.7315 10 10 m3
stream water volume in a year
V₂ = Ф₁ t
t = 1 year (365 days / year) (24 h / 1 day) (3600 s / 1h) = 3.1536 10⁷ s
V₂ = 10.1019 3.1536 10⁷
v₂ = 31,857 10⁷ m³
The total water volume is
V_win = V₁ + V₂
V_win = 6.7315 10¹⁰ + 31.857 10⁷
V_win = 7.05 10¹⁰ m³
2) let's find the amount of water lost from the lake
volume of water to surrounding bodies
V₃ = Ф₂ t
V₃ = 20.310 3.1536 10⁷
V₃ = 6.40496 10 8 m3
Volume of evaporated water
V₄ = A h₂ t
V₄ = 82 100 10⁶ 18.1 10⁻³ 12
V⁴ = 1,783 10¹⁰ m³
groundwater volume
V⁵ = 7.83.33 10³ m³
The volume of water lost is
V_lost = V₃ + V₄ + V₅
V_ lost = 6.40496 10⁸ + 1.783 10¹⁰ + 783.33 10³
V_ lost = 1.8471 10¹⁰ m³
3) the change in the height of the lake
the change in volume is
ΔV = V_ won - V_ lost
ΔV = 7.05 10¹⁰ - 1.8471 10¹⁰
ΔV = 5.20 10¹⁰ m³
the volume is
v = A h
h = V / A
h = 5.20 10¹⁰/82100 10⁶
h = 6.334 10⁻¹ m
h = 633.4 mm
Which of the following is true about how the universe is expanding?
A. galaxies are getting bigger
B. the space between galaxies is getting bigger
C. the edge of the universe is expanding, but the middle isn't.
Answer:
I'm pretty sure its b
Explanation:
A large gas-turbine power plant delivers a net power output of 325 MW to an electric generator. The minimum temperature in the cycle is 300 K, and the maximum temperature is 1500 K. The minimum pressure in the cycle is 100 kPa, and the compressor pressure ratio is 12. Calculate the power output of the turbine, the back work ratio, and the thermal efficiency of the cycle. Also, determine the improved thermal efficiency if a regenerator, with an effectiveness of 85 percent, were installed in the power plant.
Answer:
A) Power output = 541.67 MW
B) Backwork ratio = 0.4
C) Improved thermal efficiency = 0.598
Explanation:
We are given;
Net power output; W'_net = 325 MW
Minimum Temperature; T1 = 300 K
Maximum Temperature; T4 = 1500 K
Compression Pressure ratio;P2/P1 = 12
From online tables, we have the following properties of air;
Specific heat capacity; C_p = 1.004 KJ/Kg
Adiabatic constant; k = 1.4
Temperature at stage 2 will be given by the formula;
T2 = T1(P2/P1)^((k - 1)/k)
Plugging in the relevant values gives;
T2 = 300(12)^((1.4 - 1)/1.4)
T2 = 610.18 K
Similarly, Temperature at stage 3 will be;
T3 = T4(P2/P1)^((k - 1)/k)
Plugging in the relevant values gives;
T3 = 750(12)^((1.4 - 1)/1.4)
T3 = 1525.45 K
Now, let's calculate the specific heat addition given by the formula;
q = C_p(T3 - T2)
q = 1.004(1525.45 - 610.18)
q = 918.93 KJ/Kg
Let's now calculate the specific net work output;
w_net = C_p[(T3 - T4) - (T2 - T1)]
w_net = 1.004[(1525.45 - 750) - (610.18 - 300)]
w_net = 467.13 KJ/Kg
A) Power output is given by;
W_T = (W'_net/W_net) × C_p(T3 - T4)
W_T = (325/467.13) × 1.004(1525.45 - 750)
W_T = 541.67 MW
B) back work ratio is;
f = (T2 - T1)/(T3 - T4)
f = (610.18 - 300)/(1525.45 - 750)
f = 0.4
C) Thermal efficiency is given by;
η = w_net/q
Since we are told that a regenerator, with an effectiveness of 85 percent, were installed in the power plant.
Thus;
η = (w_net/q)÷0.85
η = (467.13/918.94) ÷ 0.85
η = 0.598
g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultimate strength before and after the cold-work operation. Determine the percent increase in each strength. (b) Determine the ratios of ultimate strength to yield strength before and after the cold-work operation. What does the result indicate about the change of ductility of the part
Answer:
A) - Yield strength before operation = 32 kpsi
- Ultimate Strength before operation = 49.5 kpsi
- Yield strength after operation = 61.854 kpsi
- Ultimate Strength after operation = 61.875 kpsi
- Percentage increase of yield strength = 93.29%
- Percentage increase of ultimate strength = 25%
B) ratio before operation = 1.55
Ratio after operation = 1
Explanation:
From online values of the properties of this material, we have;
Yield strength; S_y = 32 kpsi
Ultimate Strength; S_u = 49.5 kpsi
Modulus; m = 0.25
Percentage of cold work; W_c = 0.2
S_o = 90 kpsi
A) Let's calculate the strain(ε) from the formula;
A_o/A = 1/(1 - W_c)
A_o/A = 1/(1 - 0.2)
A_o/A = 1.25
Thus, strain is;
ε = In(A_o/A)
ε = In(1.25)
ε = 0.2231
Yield strength after the cold work operation is;
S'_y = S_o(ε)^(m)
Plugging in the relevant values;
S'_y = 90(0.2231)^(0.25)
S'_y = 61.854 kpsi
Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%
Ultimate strength after the cold work operation is;
S'_u = S_u/(1 - W_c)
S'_u = 49.5/(1 - 0.2)
S'u = 61.875 kpsi
Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%
B) Ratio of ultimate strength and yield strength before cold work operations is;
S_u/S_y = 49.5/32
S_u/S_y = 1.547
Ratio of ultimate strength and yield strength after cold work operations is;
S'_u/S'_y = 61.875/61.854 = 1
The ratio after the operation is less than before the operation, thus the ductility reduced.
Transmission cleaners are used: A) Only in conjunction with fuel system cleanersB) Only in the colder monthsC) By themselvesD) In conjunction with a transmission fluid exchange
Answer: D
Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.
Transmission cleaners are used in conjunction with a transmission fluid exchange. The correct option is D.
What are transmission cleaners?An exclusive additional cleaning package is included with the transmission cleaner. It pushes the deposits into the oil tray ground after removing them from the transmission.
The deposits are removed from the transmission with the regular transmission oil during an oil change. A smooth-shifting performance and a longer transmission lift time are guaranteed by the use of the transmission cleaner.
You replace your filter when you change your transmission fluid. You clean the area where the filter was when you do that. Then you can put the new filter and the new fluid on.
Therefore, in addition to changing the transmission fluid, transmission cleaners are employed. The right answer is D.
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A standard wheel consists of the ___ and the ___.
Answer:
metal hub and the wire tension spokes?
Answer: metal hub and the wire tension spokes is correct
What is a maintenance sheet?
A cook uses a pair of twelve-inch-long tongs to remove a piece of chicken from the grill. The chef is applying 3 pounds of squeezing force to the tongs. If more than 1 pound of force is applied to the fish, it will break.
The AMA of the system (rounded to the nearest hundredth) is
.
Using the static equilibrium calculations, the tongs must be held
inches (rounded to the nearest whole number) away from the fulcrum to avoid damaging the fish.
Answer: yes because fish its like quick sand is will just fall apart into tiny pecies to now it want us to round to the nearst hundredth so what is 12 divide by 3 and time 1
12:3x1=4 because we are rounding to the nearst hundredth
=4
Explanation:
Ammonia is contained in a closed, rigid 5-m3 tank initially a two-phase liquid-vapor mixture with a pressure of 2 bar. The tank contents are heated and at the final state the pressure is 4 bar and the specific volume is 0.39550 m3/kg. The effects of kinetic and potential energy are negligible. For the ammonia as the closed system, determine the amount of energy transfer by heat, in kJ per kg of ammonia.
Answer:
[tex]619.917 kJ/kg[/tex]
Explanation:
As the tank is closed and rigid, so, the mass inside the tank is fixed and the volume, [tex]V[/tex], remains constant which is
[tex]V=5 m^3[/tex]
For the pressure, P=2 bar, the thermodynamic state of ammonia on the boiling curve are
The specific volume of the liquid ammonia, [tex]v_f=0.0015 m^3/kg[/tex]
The specific volume of the vapor ammonia, [tex]v_g=0.595 m^3/kg[/tex]
The specific enthalpy of the liquid ammonia, [tex]h_f=113.464 kJ/kg[/tex]
The specific enthalpy of the vapor ammonia, [tex]h_g=1439.291 kJ/kg[/tex]
At pressure, P=4 bar
The given specific volume, [tex]v= 0.39550 m^3/kg[/tex].
Let [tex]v_1[/tex] and [tex]v_2[/tex] be the specific volumes of Ammonia in the tank. As the tank is closed, so there is no flow of Ammonia to or from the tank. Hence the mass remains constant. Also, the volume of the tank is constant, so the specific volume of Ammonia in the tank remains constant.
Hence, [tex]v_1=v_2=0.39550 m^3/kg[/tex]
Let [tex]x_1[/tex] be the vapor quality for initial condition, so
[tex]x_1=\frac{v_1-v_f}{v_g-v_f}[/tex]
[tex]\Rightarrow x_1= \frac{0.39550-0.0015}{0.595-0.0015}=0.664[/tex]
So, the initial specific energy, the liquid-vapor mixture of Ammonia have, is
[tex]e_1=h_f+x_1(h_g-h_f)[/tex]
[tex]\Rightarrow e_1=113.464+0.664(1439.291-113.464)=993.813 kJ/kg[/tex]
Now, at pressure, P=4 bar and [tex]v_2=0.39550 m^3/kg[/tex], total amount exist in the vapor phase having the specific enthalpy [tex]1613.73 kJ/kg[/tex]
So, the final energy, Ammonia have, is
[tex]e_2=1613.73 kJ/kg[/tex]
So, change in energy per kilogram of Ammonia is
[tex]\Delta e=e_2-e_1=1613.73-993.813=619.917 kJ/kg[/tex].
As the change of kinetic and potential energy are negligible and work done by the system (Ammonia in the tank) is zero due to the rigid tank, so the energy added to the tank can cause only change in energy of the system (Ammonia in the tank).
Hence, the amount of energy transfer by heat
[tex]= \Delta e= 619.917 kJ/kg[/tex].
PLEASE HELP!!!! You are going to create a transcript for a persuasive presentation to convince your client that your design improvement is the best possible solution for their needs. Take your client through the different design options you considered for the product design improvement. Explain why you discarded the designs that were not the best possible solution. Finally, explain why your final product design covers each of the requirements and specifications you used to create your needs statement.
When we drive our car at 90 feet per second [ft/s], we measure an aerodynamic force (called drag) of 59 pounds-force [lbf] that opposes the motion of the car. How much horsepower [hp] is required to overcome this drag?
Answer:
P = 9.65 hp
Explanation:
Given that,
Speed of a car, v = 90 ft/s
Force that opposes the motion of the car, F = 59 pounds-force
90 ft/s = 27.432 m/s
1 pound-force = 4.448 N
59 pounds-force = 262.445 N
Power required is given by :
[tex]P=F\times v\\\\P=262.445\ N\times 27.432\ m/s\\\\P=7199.391\ W[/tex]
Since, 1 hp = 745.7
7199.391 W = 9.65 hp
So, 9.65 hp of power is required to overcome the drag.
Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm and r2 = 6.5 cm, respectively. The outer surface of the pipe is wrapped with a thin electric heater that consumes 300 W per m length of the pipe. The exposed surface of the heater is heavily insulated so that all heat generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection with a heat transfer coefficient of h = 85 W/m2⋅K. Assume that the thermal conductivity is constant and the heat transfer is one-dimensional. Express the mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in the pipe during steady operation.
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffered from a torque (T) of 100 Nmat one end (fixed at the other end). By neglecting the outside pressureand radial stresses, choose a proper 2D element and showcase the non-zero stresses on the element. Then, determine all three principal stresses and the maximum shear stress at that point.
Answer:
All 3 principal stress
1. 56.301mpa
2. 28.07mpa
3. 0mpa
Maximum shear stress = 14.116mpa
Explanation:
di = 75 = 0.075
wall thickness = 0.1 = 0.0001
internal pressure pi = 150 kpa = 150 x 10³
torque t = 100 Nm
finding all values
∂1 = 150x10³x0.075/2x0,0001
= 0.5625 = 56.25mpa
∂2 = 150x10³x75/4x0.1
= 28.12mpa
T = 16x100/(πx75x10³)²
∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]
= 1/2[84.37±√791.2969+5.827396]
= 1/2[84.37±28.33]
∂1 = 1/2[84.37+28.33]
= 56.301mpa
∂2 = 1/2[84.37-28.33]
= 28.07mpa
This is a 2 d diagram donut is analyzed in 2 direction.
So ∂3 = 0mpa
∂max = 56.301-28.07/2
= 14.116mpa
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb. (Jr., 01/2018, p. P-5) Jr., W. C., Rethwisch, D. G. (2018). Materials Science and Engineering: An Introduction, Enhanced eText, 10th Edition. [[VitalSource Bookshelf version]]. Retrieved from vbk://9781119405498 Always check citation for accuracy before use.
Answer:
The theoretical density of the Niobium is 8.57 g/cm³.
Explanation:
For a BCC crystal structure the density can be calculated using the following equation:
[tex] \rho = \frac{nA}{VN} [/tex]
Where:
n: is the number of atoms, for a BCC = 2 atoms
A: is the atomic weight = 92.91 g/mol
V: is the unit cell's volume
N: is the Avogadro's number = 6.022x10²³ atom/mol
The volume of the unit cell of a BCC crystal structure (V) can be found as follows:
[tex]V = (\frac{4R}{\sqrt{3}})^{3}[/tex]
Where R is the atomic radius = 0.143 nm
[tex]V = ({\frac{4R}{\sqrt{3}}})^{3} = ({\frac{4*0.143 nm*\frac{1 cm}{1\cdot 10^{7} nm}}{\sqrt{3}}})^{3} = 3.60 \cdot 10^{-23} cm^{3}[/tex]
Now, the density is:
[tex] \rho = \frac{nA}{VN} = \frac{2 atom*92.91 g/mol}{3.60 \cdot 10^{-23} cm^{3}*6.022 \cdot 10^{23} atom/mol} = 8.57 g/cm^{3} [/tex]
Therefore, the theoretical density of the Niobium is 8.57 g/cm³.
I hope it helps you!
A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and the return activated sludge concentration is 8000 mg/L. The biomass concentration is 3500 mg/L. The combined design flow rate of the primary and secondary clarifiers is 2.5 MGD. After primary treatment, the wastewater has an influent BOD concentration of 200 mg/L and an influent suspended solids concentration of 200 mg/L. Two secondary clarifiers, each 28 ft in diameter, are then used. After secondary treatment, the effluent BOD concentration is 15 mg/L, and the effluent suspended solids concentration is 20 mg/L. The volume of sludge produced is 0.5 MGD. What is most nearly the solids residence time
Answer:
11 hours approximately
Explanation:
We are to calculate mean cell residence time mcrt
= Mass of solid in reactor/mass of solid wasted in a day
Q = Qe + We
Q = 2.5
Qw = 0.5
Qe = 2.5 - 0.5
= 2 MGD
10⁶/svi
= 10⁶/125
= 8000
X = 3500
Xe = 20mg/
1MGD = 0.1337million
Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337
= 262500000/[4000000000+40000000} x 0.1337
= 262500000/574800000
= 0.45668 days
= 0.45668 x 24 hours
= 10.9603 hours
Approximately 11 hours
There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W). Monitor a cellular phone call and observe the type of telephone and the speed of the user. The probability model for this experiment has the following information:
P[F]=0.5, P[HF]=0.2, P[MW]=0.1.
What is the sample space of the experiment?
Find the following probabilities P[W], P[MF], and P[H].
Answer:
A) P(W) = 0.5
B) P(MF) = 0.3
C) P(H) = 0.6
Explanation:
We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.
Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).
Thus, the sample space is combination of types and classification we are given and it is written as;
S = {HF, HW, MF, MW}
A) Now, phones can either be fast(F) or slow(W). Thus, we can write;
P(F) + P(W) = 1
We are given P(F) = 0.5
Thus;
0.5 + P(W) = 1
P(W) = 1 - 0.5
P(W) = 0.5
B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;
P(H ∩ F) + P(M ∩ F) = P(F)
We are given P[F] = 0.5 and P[HF] = 0.2.
P(H ∩ F) is same as P[HF]
Also, P(M ∩ F) is same as P(MF)
Thus;
0.2 + P(MF) = 0.5
P(MF) = 0.5 - 0.2
P(MF) = 0.3
C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}
Thus;
P(M) = P(MW) + P(MF)
P(M) = 0.1 + 0.3
P(M) = 0.4
Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;
P(H) + P(M) = 1
P(H) + 0.4 = 1
P(H) = 1 - 0.4
P(H) = 0.6
Tech A says that horsepower is a measurement simply of the amount of work being performed. Tech B says that horsepower can be calculated by multiplying torque by rpm and dividing by 5252. Who is correct?
Answer:
Tech B
Explanation:
Horsepower (hp) refers to a unit of measurement of power in respect of the output of engines or motors.
Horsepower is the common unit of power. It indicates the rate at which work is done.
The formula [tex]\frac{rpm*T}{5252}[/tex], where rpm is the engine speed, T is the torque, and 5,252 is radians per second.
So,
Tech B is correct
The evaporator section of a refrigeration unit consists of thin-walled, 10-mm-diameter tubes through which refrigerant passes at a temperature of −18°C. Air is cooled as it fows over the tubes, maintaining a surface convection coeffcient of 100 W/m2 ⋅ K, and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of −3°C, what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator’s defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range 0 ≤δ ≤ 4 mm. Frost may be assumed to have a thermal conductivity of 0.4 W/m ⋅ K. (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which T[infinity] = 20°C and natural convection maintains a convection coeffcient of 2 W/m2 ⋅ K, how long will it take for the frost to melt? The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of fusion of 334 kJ/kg
A wind turbine-electric generator is mounted atop a tower. As wind blows steadily across the turbine blades, electricity is generated. The electrical output of the generator is fed to a storage battery.
(a) Considering only the wind turbine-electric generator as the system, identify locations on the system boundary where the system interacts with the surroundings.
(b) Repeat for a system that includes only the storage battery.
What is the purpose of a career portfolio?
to introduce a resume and ask for a job interview
to provide relevant information in a job application
to keep track of all of the jobs you’ve applied for throughout your career
to collect work samples and the data that is used in a resume
Answer:
c
Explanation:
Answer: The Answer is B
Explanation:
A parallel helical gearset consists of a 19-tooth pinion driving a 57-tooth gear. The pinion has a left-hand helix angle of 30°, a normal pressure angle of 20°, and a normal module of 2.5 mm.
Find:_______.
(a) The normal, transverse, and axial circular pitches
(b) The transverse diametral pitch and the transverse pressure angle
(c) The addendum, dedendum, and pitch diameter of each gear
Answer:
a)
normal circular pitch = 7.8539 mm
transverse circular pitch = 9.0689 mm
axial circular pitches = 15.7077
b)
transverse diametral pitch is 0.3464 teeth/mm
transverse pressure angle is 22.8°
c)
Addendum = 2.5 mm
dedendum = 3.125 mm
pinion diameter = 54.8482 mm and Gear diameter = 164.5448 mm
Explanation:
Given that;
module m = 2.5 mm
Number of teeth on Gear nG = 57 TEETH
Number of teeth on Pinion nP = 19 TEETH
Helix angle W = 30°
Normal Pressure angle β = 20°
finding the circular pitch
Pc = πm
we substitute
Pc = π * 2.5 mm = 7.8539 mm
now the diametral pitch p = π / Pc
= π / 7.8539
= 0.4 teeth/mm
a)
So the normal circular pitch
Pn = π / P
Pn = π / 0.4
Pn = 7.8539 mm
the transverse circular pitch
Pt = Pn / cosW
Pt = 7.8539 / cos30°
Pt = 9.0689 mm
for axial circular pitches
Px = Pt / tanW
Px = 9.0689 / tan30°
Px = 15.7077
b)
The transverse diametral pitch and the transverse pressure angle.
The transverse diametral pitch Pt = PcosW
= 0.4 * cos30°
= 0.3464 teeth/mm
transverse diametral pitch is 0.3464 teeth/mm
transverse pressure angle β1 = tan^-1 ( tan βn / cos W)
= tan^-1 ( tan20° / cos 30°)
= tan^-1 ( 0.42027 )
β1 = 22.8°
transverse pressure angle is 22.8°
c)
The addendum, dedendum, and pitch diameter of each gear
Now from table standard Tooth proportions for Helical Gears;
Addendum a = 1/p
= 1 / 0.4
= 2.5 mm
dedendum b = 1.25 / p
= 1.25 / 0.4
= 3.125 mm
now pinion diameter dP = Np / PcosW
= 19 / 0.4 (cos30°)
= 54.8482 mm
Gear diameter dG = nG / pcosW
= 57 / 0.4 (cos30°)
= 164.5448 mm
Which conditions contribute to engine deposits
Driving consistently will cause the oil to heat up and break down to create sludge.
And if oil gets too hot, it will cause metal on metal grinding due to the oil thinning out.
Hope this helps you. :)
A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 step-down transformer having a single secondary winding. It uses four diodes, each of which can be modeled to have a 0.7-V drop for any current. What is the peak value of the rectified voltage across the load? For what fraction of a cycle does each diode conduct? What is the average voltage across the load? What is the average current through the load?
Answer:
The average current will be "10.12 mA".
Explanation:
In the transformer:
⇒ [tex]\frac{V1}{V2}=\frac{N1}{N2}[/tex]
where,
V1 = 120
N2 = 1
N1 = 10
So that,
⇒ [tex]V2=V1\times \frac{N2}{N1}[/tex]
On putting the values,
[tex]=120\times \frac{1}{10}[/tex]
[tex]=12V \ rms[/tex]
The full wave rectifier would conducts during positive and negative half.
⇒ [tex]Vmax=12-0.7[/tex]
[tex]=11.3V \ rms[/tex]
⇒ [tex]Vpeak=11.3\sqrt{2}[/tex]
[tex]=15.9V[/tex]
⇒ [tex]Vaverage=2Vmax \ \pi[/tex]
[tex]=10.12 \ V[/tex]
The amount conducted by each diode is 50%.
⇒ [tex]Iaverage=\frac{Vaverage}{R}[/tex]
[tex]=\frac{10.12}{1}[/tex]
[tex]=10.12 \ mA[/tex]
A new car design must carry a family of four, include safety features, get more than 25 miles per gallon on the highway, and weigh less than 1.5 tons. which is one of the criteria that should be part of the design to meet the proposal request?
Answer: weigh less than 1.5 tons
Explanation:
Answer:The answer is A, seats for five people.
Explanation: I took the test and it is the only option that fits the criteria listed above.
