Answer:
b
Explanation:
An ignition coil is an example of a. A. Step-up transformer. B. Step- down transformer. C. Relay.
D. Solenoid.
Answer:
D
Explanation:
An ignition coil is an example of a Solenoid. Thus, the correct option for this question is D.
What is Solenoid?Solenoids may be characterized as a coil of wire that is usually in a cylindrical form and carries a current that acts like a magnet. Due to this, a migratory core is drawn into the coil when a current flows, and that is utilized especially as a switch or control for a mechanical device.
Similarly, an ignition coil also represents an example of a solenoid because it also consists of a coil of wire and carries a current that acts like a magnet. While the concept of the transformers is different as it stores a huge amount of electric current and spread it accordingly.
Therefore, an ignition coil is an example of a Solenoid. Thus, the correct option for this question is D.
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For many clients, a decision is based primarily on time and money, which are the two most abundant
resources a company has.
True or False
In order to avoid slipping in the shop, your footwear should ___________.
(This is for my automotive class by the way)
Answer:
good shoes
Explanation:
yes
list the major parts in a shop's air supply system
Answer:
Compressed air systems usually consist of the following components:
compressor.
air cooler.
air receiver tank.
filter.
dryer.
condensate trap.
distribution system
Answer:
Compressed air systems usually consist of the following components:
compressor.
air cooler.
air receiver tank.
filter.
dryer.
condensate trap.
distribution system
Explanation:
Compressed air systems usually consist of the following components:
compressor.
air cooler.
air receiver tank.
filter.
dryer.
condensate trap.
distribution system
who designs each page or screen with which a user interacts?
Answer:
UI Design
Explanation:
(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom. Air flows from the tube into the liquid and creates spherical bubbles with diameter about the diameter of the tube (1 mm). Everything is at 298 K. The tube is short but is connected to a much longer 2 m long hose that is 6 mm in diameter. The hose is connected to the gas supply. If there is no gas flow the water will leak into the tube and into the supply hose. When gas flows the water is blocked from entering the tube and bubbling starts. State all assumptions in answering the following questions. (a) What should be the minimum air flow rate and the gas supply pressure to keep the water from leaking back into the tube? (b) Is the flow in the hose laminar or turbulent? Is the flow in the tube laminar or turbulent?
Solution :
Given :
h = 2 cm
Diameter of the tube , d = 1 mm
Diameter of the hose, D = 6 mm
Between 1 and 2, by applying Bernoulli's principle, we get
As point 1 is just below the free surface of liquid, so
[tex]$P_1=P_{atm} \text{ and} \ V_1=0$[/tex]
[tex]$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$[/tex]
[tex]$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$[/tex]
[tex]$P_2 = 111.35 \ kPa$[/tex]
Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.
Velocity at point 2,
[tex]$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$[/tex]
= 1.617 m/s
Flow of water, [tex]$Q_2 = A_{tube} \times V_2$[/tex]
[tex]$=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617 $[/tex]
[tex]$1.2695 \times 10^{-6} \ m^3/s$[/tex]
Minimum air flow rate,
[tex]$Q_2 = Q_3 = A_{hose} \times V_3$[/tex]
[tex]$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$[/tex]
[tex]$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$[/tex]
= 0.0449 m/s
b). Reynolds number in hose,
[tex]$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$[/tex]
υ for water at 25 degree Celsius is [tex]$8.9 \times 10^{-7} \ m^2/s$[/tex]
υ for air at 25 degree Celsius is [tex]$1.562 \times 10^{-5} \ m^2/s$[/tex]
[tex]$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$[/tex]
= 17.25
Therefore the flow is laminar.
Reynolds number in the pipe
[tex]$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$[/tex]
= 1816.85, which is less than 2000.
So the flow is laminar inside the tube.
The user manual states an accuracy of 0.1% of the reading for a tachometer (an instrument measuring the rotation speed of a shaft or disk, as in a motor or other machine) with an analog display dial graduated in 5 revolutions per minute (rpm) increments. Estimate the design stage uncertainty at 5000 rpm.
Answer: the design stage uncertainty is ±5.6 rpm
Explanation:
Given that;
machine with an analog display dial graduated in 5 revolutions per minute (rpm) increments; Q = 5 rpm
now lets calculate zero order uncertainty in this machine
⇒ ± 1/2Q = 1/2(5) = ± 2.5 rpm
given that accuracy is 0.1% of reading
0.1% of 5000 rmp = 0.1/100 × 5000 = 5 rpm
now the design stage uncertainty at 5000 rpm will be;
⇒ √((5)² + (2.5)²)
= √(25 + 6.25)
= √ 31.25
= ±5.59 ≈ ±5.6 rpm
Therefore the design stage uncertainty is ±5.6 rpm
You are designing a critical component to be installed in a chemical plant, where corrosion resistance and high temperature performance are critical. As a result, a fully dense material is required. Because of the materials used in the system, the part can only be produced by powder metallurgy. Which powder metallurgy process would you use to produce a fully dense component?a) Hot Isostatic Pressing b) Sinter then Metal Injection Molding c) Cold Isostatic Pressing d) Press and Sinter Processes e) Metal Injection Molding
Answer:
Yes
Explanation:
Because your installing something
The powder metallurgy process that would be used to produce a fully dense component is Hot Isostatic Pressing. The correct option is a.
What is Hot Isostatic Pressing?Hot isostatic pressing is a process used to consolidate metal powder or to remove defects in solids such as pores, voids, and internal cracks, densifying the material to 100% of its theoretical density.
Hot Isostatic Pressing (HIP) is a process used to densify powders, cast and sintered parts, and steels and superalloys in a furnace at high pressure and temperatures ranging from 900 to 1250°C.
To provide isotropic properties and 100% densification, the gas pressure acts uniformly in all directions.
Hot Isostatic Pressing is the powder metallurgy process that would be used to create a fully dense component.
Thus, the correct option is a.
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Select three mechanical processes used in the manufacturing phase of engineering.
Drying
Mixing
Cooling
Crystallization
Size reduction
Food packaging
Answer:Size reduction
Cooling
Mixing
Explanation:
Which of these credit building options do you personally think is the easiest method that you can see yourself doing? Explain your reasoning.
Answer:
Explanation:
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The credit building options that I think is the easiest method that you can see yourself doing is Secured credit cards.
What is the Secured credit cards?Secured credit cards are known to be a type of card that are said to be backed by a specific refundable security deposit.
This kind of deposit is known to limit or lower the lender's risk and thus makes it better for people who do not have credit or when they do not qualify for that type of credit. This credit option is easier to get when compared to unsecured cards, as it does not need a lot of requirement.
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Electricians will sometimes call ______ "disconnects" or a "disconnecting means."
A) Two-pronged plugs
B) Electrical cords
C) Switches
D) Three-pronged plugs
(This is for my Automotive class)
Answer:
C.switches
Explanation:
30 KJ of heat energy is supplied to a 4.5 kg block of aluminum at 20°C.Calculate the final temperature of the block,if the specific heat capacity of aluminum is 948 J kg^-1 K^-1 and no loss of heat energy
What is the step in which you are testing your hypothesis
Answer:
Step 1: State your null and alternate hypothesis. ...
Step 2: Collect data. ...
Step 3: Perform a statistical test. ...
Step 4: Decide whether the null hypothesis is supported or refuted. ...
Step 5: Present your findings.
_____ refers to the knowledge and skills necessary to manage, control, and work with hazardous materials and pollutants.
A bronze bushing 60 mm in outer diameter and 40 mm in inner diameter is to be pressed into a hollow steel cylinder of 120-mm outer diameter. Calculate the tangential stresses for steel and bronze at the boundary between the two parts
Answer:
The tangential stress for the steel at 30mm = 88.8MPa.
The tangential stress for the steel at 60mm = 35.5 MPa.
The tangential stress for the bronze at 20mm = - 191.9 MPa.
The tangential stress for the bronze at 30mm = - 138.6 MPa.
Explanation:
The outer radius bronze bushing = 60 mm/2 = 30 mm, the inner radius for bronze bushing = 40mm/2 = 20mm and the cylinder radius = 120mm/2 = 60mm.
Step one: The first step is to calculate or Determine the interface pressure.
The interface pressure = 0.05/ [ [ 30 × ( 1/210 ×10⁹) ] × [( 60² + 30²/ ( 60² - 30²) + 0.3 ] + [ 1/105 × 10⁹× [ ( 20² + 30²)/( 30² - 20²) - 0.3] ].
The interface pressure = 53.3 MPa
Step two: Determine the tangential stresses for steel and bronze as given below:
The tangential stress for the steel at 30mm = 10⁶ × 53.3 [ ( 0.06)² + (0.03)²/ ( 0.06)² - (0.03)² ] = 88.8 MPa.
The tangential stress for the steel at 60mm = 10⁶ × 53.3 × 2 × [0.03]²/ ( 0.06)² - (0.03)² ] = 35.5 MPa.
The tangential stresses for the bronze is calculated below;
=> The tangential stress for the bronze at 20mm = - [ 10⁶ × 53.3 × 2 × [0.03]² ] / ( 0.03)² - (0.02)² ] = - 191.9 MPa.
The tangential stress for the bronze at 30mm = - [ 10⁶ × 53.3 × [ 0.03]² + (0.03)² ] / ( 0.03)² - (0.02)² ] = - 138.6 MPa.
Water vapor at a rate of 40,000 kg/h and at 8 MPa and 500 C enters an adiabatic turbine and leaves the turbine at 40 kPa as saturated vapor. The rate of entropy generation for this turbine is close to,
Answer:
M(dot)= 40000 kg/h Pressure= 8Mpa and T= 500C If its adiabatic then you know that it loses no heat or Q(dot) =0 You have a turbine which drives a shaft producing work for the other systems.We want to find entropy production of alpha.Naomi needs to test an application that will be accessed by multiple users simultaneously. She is planning to use LoadRunner to test the performance of the application. How can a testing tool such as LoadRunner help her with performance testing?
A.
by using a large number of testers
B.
by increasing the number of test cases
C.
by automating your test cases
D.
by creating a number of virtual users
A hubbing press has capasity of 175tons. If the workpice ia a copper alloy part with a 2in^2 projected area, what is the strongest alloy that can be hubbed?
Answer:
The answer is "[tex]\bold{175000 \ \frac{lbs}{in^2}}[/tex]"
Explanation:
The formula for the max value of UTS:
[tex]= \frac{F}{A} \\\\= \frac{175\ ton}{ 2 \ in^2} \\\\= \frac{350000}{2}\\\\[/tex]
Max UTS:
[tex]\to 175000 \ \frac{lbs}{in^2}[/tex]
An alternating current is supplied to an electronic component with a rating that it be used only for voltages below 12 V. What is the highest Vrms that can be supplied to this component while staying below the voltage limit
Answer:
[tex]V_{rms}=6\sqrt2\ V[/tex]
Explanation:
Given that,
The maximum voltage of an alternating current, [tex]V_{max}=12\ V[/tex]
We need to find the highest Vrms that can be supplied to this component while staying below the voltage limit.
Let rms voltage in terms of peak voltage is given by :
[tex]V_{rms}=\dfrac{V}{\sqrt2}\\\\=\dfrac{12}{\sqrt2}\\\\=\dfrac{12}{\sqrt2}\times \dfrac{\sqrt2}{\sqrt2}\\\\=\dfrac{12\sqrt2}{2}\\\\=6\sqrt2\ V[/tex]
Hence, the required rms voltage is [tex]6\sqrt2\ V[/tex].
Whose responsibility is it to identify confined spaces and reduce hazards? A The employee B The rescuer C The employer D OSHA
Answer:
C. The Employer
Explanation:
Under OSHA's Confined Space Standard, it is the responsibility of the "employer" to identify or evaluate the spaces whether they are "permit spaces" or "confined spaces." These spaces have serious safety hazards or health hazards. Examples: tanks, vessels, pits, underground vaults, and the like.
The employers need to disclose the information of the permit spaces and their risks to the employees. Thus, he should have a written permit of the permit space if he will be allowing the employees to enter. It is also the employer's responsibility to reduce the hazards imposed to the employees such as verifying the entry conditions that are acceptable, making sure the permit space is isolated, providing barriers, etc.
A fluid flows though a horizontal 0.1 inch diameter pipe. When the Reynold number is 1508, the head loss over a 20-ft length of the pipe is 6.2 ft. Determine the fluid velocity.
Answer:
The fluid velocity V = 1.98 ft/s
Explanation:
From the information given:
The fluid velocity can be determined from the head-loss [tex]h_L[/tex] of a laminar pipe and it is expressed as:
[tex]h_L = f\dfrac{l\times V^2}{D \times 2g}[/tex]
where;
f = frictional factor ; l = length; D = diameter; V= fluid velocity and g = acceleration due to gravity.
And;
[tex]f = \dfrac{64}{Re}[/tex]
For fluid movement in a laminar flow, the Reynolds number (Re) is usually lesser than 2100.
Given that:
Re = 1508 < 2100 ( laminar flow)
Then;
[tex]f = \dfrac{64}{1508}[/tex]
f = 0.04244
Also;
the head-loss [tex]h_L[/tex] = 6.2 ft
frictional force f = 0.04244
length = 20-ft
acceleration due to gravity (g) = 32.2 ft/s²
Replacing all the values into the equation [tex]h_L = f\dfrac{l\times V^2}{D \times 2g}[/tex]; the fluid velocity is:
[tex]6.2 = 0.04244 \times \dfrac{20 \times V^2}{0.1 \times \dfrac{1}{12} \times 2\times 32.2}[/tex]
[tex]6.2 = 0.04244 \times \dfrac{20 \times V^2}{0.53667}[/tex]
6.2 × 0.53667 = 0.04244 × 20 × V²
3.327354 = 0.8488 × V²
[tex]V^2= \dfrac{3.327354} { 0.8488}[/tex]
[tex]V^2=3.92[/tex]
[tex]V = \sqrt{3.92}[/tex]
V = 1.98 ft/s
"Well, this is a fine how-do-you-do. I come all the way over here to introduce my friends, R.W. and Melvin, to y'all and y'all actin' like y'all ain't got no manners at all. Yeah, ole R.W. and Melvin," he said, rolling the Simmses' names slowly off his tongue to bring to our attention that he had not bothered to place a "Mister" before either, "they been mighty fine friends to me. Better than any of y'all. Look, see here what they give me." Proudly he tugged at the suit coat. "Pretty nice stuff, eh? Everything I want they give me 'cause they really likes me. I'm they best friend." —Roll of Thunder, Hear My Cry, Mildred D. Taylor What do T. J.’s actions indicate?
A. He dislikes that his old friends never bought him things.
B. He is showing off his new friends’ generosity.
C. He feels sorry that his old friends do not have nice things.
D. He is angry about the way his old friends treated him.
Answer:
B. He is showing off his new friends’ generosity.
Explanation:
Answer: B He is showing off his new friends' generosity
Explanation: 2020 EDGE is correct with B
calculate the touque developed by a motor whose spindle is rotating at 1500 rev/min and developing a power of 1.80W
Answer:
n=1500/60 to get , it into rev/sec
W=2πn = 50π
T=P*W=1.8 *50π=282.74 N.M
Explanation:
Steam enters a steady-flow adiabatic nozzle with a low inlet velocity (assume ~0 m/s) as a saturated vapor at 6 MPa and expands to 1.2 MPa.
a. Under what conditions is the velocity from this nozzle maximized?
b. Determine the maximum exit velocity of the steam, in m/s.
This happens as a result of a gain in kinetic energy at the expense of a decrease in enthalpy brought on by gas expansion as air passes through the diverging region of the nozzle.
What velocity from this adiabatic nozzle maximized?A nozzle is a tool used to change the characteristics of a fluid flow as it leaves a confined chamber or conduit, particularly to enhance velocity. The reciprocal of the nozzle length, pressure, and the square root of the nozzle diameter all increase air velocity.
A nozzle needs to be made to expand steam from 500 PA at 210 °C to 100 kPa at a rate of 0.1 kg/s. In order to reach critical or sonic conditions (i.e., choked flow) at its throat, the flow is accelerated when a nozzle is used.
Therefore, The steam's specific volume at the nozzle's exit is determined to be 1.4 m3/kg and its velocity at that point to be 700 m/s.
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Which of the following is the term for a sliding device that moves against or is moved by fluid pressure?
turbine
valve
piston
fluid
Answer:
piston
a sliding device that moves against or is moved by fluid pressure; usually consists of a cylindrical body within a cylindrical chamber
Explanation:
2021 EDGE I got it correct :D
Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calculate the metal removal rate (MRR)
Answer:
Explanation:
The material rate of removal is the total amount of material removed per unit time during any operation. The material rate of removal is usually given by the formula
Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where
Radial Depth of Cut, RDOC = 25 mm
Axial depth of cut, ADOC = 200 mm
Feed rate, FR = 25 mm/min
On multiplying all together, we will then have our Material Rate of Removal, which is
MRR = 25 mm * 200 mm * 25 mm/min
MRR = 125000 mm³/min
Or we convert it to cm³/min and have
MRR = 125000 mm³/min ÷ 1000
MRR = 125 cm³/min
a centrifugal pump operating at 1000 rpm is delivering 80 gpm of water at 150 psig and requires 8 hp of shaft power at the best efficieny point determine the flow rate discharge pressure required shaft power and efficiency
Answer:
Flow rate = 118.8 gpm, discharge pressure= 331.5 psig, shaft power = 26.3 hp and Efficiency = 87.5%.
Explanation:
Without mincing words, let us dive straight into the solution to the question above. So, the following parameters or data are given in the question above:
gpm = 80 gpm, shaft power = 8hp, and the pressure of 150 psig.
The flow rate =[ 80 × 1500] ÷ 100g = 118.9 gpm.
Hence, the discharge pressure = 150 × [ 1500/100]² = 331.5 psig.
Also, the required shaft power = 8 × [ 1500/100]³ = 26.3 hp.
The last part that is the efficiency can be calculated as given below:
Efficiency = [ 80 × 0.00223 × 144 × 150 ] ÷ [550 × 8 ] = 0.875 = 87.5%.
A Piper Cub airplane has a gross weight of 1700 lb, a cruising speed of 114 mph, and a wing area of 178 ft^2. Determine the lift coefficient of this airplane for these conditions.
Solution :
As the plane is flying in the air, the force of weight is pulling the plane down. This force must be equal to the force of lift which is generated from the plane's translation true the air. Therefore, the next condition must be satisfied.
[tex]$F_W= L $[/tex] ................. (i)
The lift force of the plane is given by :
[tex]$L=C_V \frac{\rho}{2} V^2 A$[/tex]
Since the weight force of the plane is given, from the condition (i), we also know the intensity of the lift force.
Therefore, we transform the lift force expression to derive the lift coefficient formula:
[tex]$C_L=\frac{2L}{\rho V^2 A}$[/tex]
[tex]$C_L=\frac{2 \times 1700}{2.38 \times 10^{-3 }\times (167.2)^2 \times 178} $[/tex]
[tex]$C_L = 0.287$[/tex]
Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calculate the metal removal rate (MRR)
Answer:
125 cm³/min
Explanation:
The material rate of removal is usually given by the formula
Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where
Radial Depth of Cut = 25 mm
Axial depth of cut = 200 mm
Feed rate = 25 mm/min
On multiplying all together, we will then have
MRR = 25 mm * 200 mm * 25 mm/min
MRR = 125000 mm³/min
Or we convert it to cm³/min and have
MRR = 125000 mm³/min ÷ 1000
MRR = 125 cm³/min
A
AMINO BORCY
Two forces of 80 N and 70 N act simultaneously at a point Find the resultant force. if the
Angle between them is 150 dgree
Answer:
[tex]F_r = 40.01N[/tex]
Explanation:
Given
Let the two forces be [tex]F_1\ and\ F_2[/tex]
[tex]F_1 = 80N[/tex]
[tex]F_2 = 70N[/tex]
[tex]\theta = 150\deg[/tex]
Required
Determine the resultant force [tex]F_r[/tex]
This will be solved using parallelogram law of resultant force.
The equation is:
[tex]F_r^2 = F_1^2 + F_2^2 + 2F_1F_2cos\theta[/tex]
Substitute values for [tex]F_1[/tex], [tex]F_2[/tex] and [tex]\theta[/tex]
[tex]F_r^2 = 80^2 + 70^2+ 2 * 70 * 80 * cos\ 150[/tex]
[tex]F_r^2 = 6400 + 4900 + 11200* cos\ 150[/tex]
[tex]F_r^2 = 6400 + 4900 + 11200* -0.8660[/tex]
[tex]F_r^2 = 6400 + 4900 - 9699.2[/tex]
[tex]F_r^2 = 1600.8[/tex]
Take the square root of both sides
[tex]F_r = \sqrt{1600.8[/tex]
[tex]F_r = 40.01N[/tex]
Hence, the resultant force is 40.01 N