Answer:
Option B is correct
Explanation:
Electron and protons represent current in opposite direction as they have opposite charges on it. Due to motion of electron and proton in electric field they will produce the magnetic field. As both are moving parallel it means they generate the magnetic field which is in the same direction as of their electrostatic force.
Option B is correct:
The moving electrons and protons represent currents in opposite directions; thus, they will exert a magnetic force that will be in the same direction of their electrostatic force.
4. Oil system cleaning products should not use solvents because:
A) O Solvents smell bad
B) Solvents are completely removed during the service
CO Solvents have no impact on dirt and debris
DO Solvents can damage certain plastics and rubbers found in the engine
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Answer:
Solvents can damage certain plastics and rubbers found in the engine
Explanation:
Can anyone tell me all the corrects answers to these? I’m sorry if this is the wrong subject I’m not sure what to put it under but I really need help!
Answer:
Crankshaft position sensor - F I can't quite make out the letter but it's the thing at the bottom almost touching the notched wheel.
Coil Module - B
Knock Sensor - D
Coil Pack -E
Fuse Block - A
Powertrain Control Module - C
Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and 50°C to a final state at which the refrigerant is saturated vapor. For the refrigerant, determine the work and heat transfer, per unit mass, each in kJ/kg. Any changes in kinetic and potential energy are negligible.
Answer:
- Work done is 2.39 kJ
- heat transfer is 20.23 kJ/kg
Explanation:
Given the data in the question;
First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;
Specific Volume v₁ = 0.02547 m³/kg
Specific enthalpy u₁ = 243.78 kJ/kg
Specific Volume V₂ = 0.02846 m³/kg
Specific enthalpy u₂ = 261.62 kJ/kg
p = 8 bar = 800 kPa
Any changes in kinetic and potential energy are negligible.
So we determine the work done by using the equation at constant pressure
]Work done W = p( v₂ - v₁ )
we substitute
W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )
W = 800 kPa( 0.00299 m³/kg )
W = 2.39 kJ
Therefore, Work done is 2.39 kJ
Heat transfer;
using equation at constant pressure
Heat transfer Q = W + ( u₂ - u₁ )
so we substitute
Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )
Q = 2.392 kJ + 17.84 kJ/kg )
Q = 20.23 kJ/kg
Therefore, heat transfer is 20.23 kJ/kg
What's a major difference between a construction drawing and a schematic diagram
Answer:
Explanation:
When used as nouns, schematic diagram means a plan, drawing, sketch or outline to show how something works, or show the relationships between the parts of a whole, whereas schematic means a simplified line-drawing generally used by engineers and technicians to describe and understand how a system works at an abstract level.
Hope this helped!!!
Guess the output of this code: print( (3**2)//2)
the answer is 2
kids simple just multiply 3 by 2 which gets the answer as 6 then divide 6 by 2 then it comes to again 2 and that is a simple answer...
The output of the given code, i.e., print( (3**2)//2) will be 4.
What is coding?Coding, also known as computer programming, is the method by which we communicate with computers.
Code tells a computer what to do, and writing code is similar to writing a set of instructions. You can tell computers what to do or how to behave much more quickly if you learn to write code.
When you choose an instructional programming language, coding is simple to learn. It can be difficult to learn to code if you begin with a more complex coding language.
The ** operator is used to multiply a number by a power.
The // operator calculates how many times the right number can fit into the left.
Here it is given that
[tex]print(3^2)/2[/tex]
So, the round division will be 4.
Thus, the output will be 4.
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The top surface of an L = 5mmthick anodized aluminum plate is irradiated with G = 1000 W/m2 while being simultaneously exposed to convection conditions characterized by h = 50 W/m2 ⋅ K and T[infinity] = 25°C. The bottom surface of the plate is insulated. For a plate temperature of 400 K as well as α = 0.14 and ε = 0.76, determine the radiosity at the top plate surface, the net radiation heat flux at the top surface, and the rate at whic
Answer:
[tex]J=1963W/m^2[/tex]
[tex]q_{rad}=963w/m^2[/tex]
[tex]\triangle T= -0.378k/s[/tex]
Explanation:
From the question we are told that:
[tex]L=5mm => 5*10^{-3}\\Irradiation G=1000W/m^2\\h=50W/m^2\\T_{infinity} = 25C.\\Plate\ temperature\ T_p= 400 K\\\alpha=0.14\\E=0.76[/tex]
at Temp=400K
[tex]E=2702kg/m^2,c=949J/kgk[/tex]
Generally the equation for Radiosity is mathematically given by
[tex]J=eG+\in E_p[/tex]
[tex]J=(1-\alpha)G+\in \sigma T^4[/tex]
[tex]J=(1-0.14)1000+0.76 (5.67*10^_{8}) (400)^4[/tex]
[tex]J=1963W/m^2[/tex]
Generally the equation for net radiation heat flux [tex]q_{rad}[/tex] is mathematically given by
[tex]q_{rad}=J-G\\q_{rad}=1963-1000[/tex]
[tex]q_{rad}=963w/m^2[/tex]
Generally the equation for and the rate of plate temp [tex]\triangleT[/tex] is mathematically given by
[tex]\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}[/tex]
[tex]\triangle T= \frac{45(400)-(30+273+963)}{(2702*949*0.005)}[/tex]
[tex]\triangle T= -0.378k/s[/tex]
A fuel oil is burned with air in a furnace. The combustion produces 813 kW of thermal energy, of which 65% is transferred as heat to boiler tubes that pass through the furnace. The combustion products pass from the furnace to a stack at 6500C. Water enters the boiler tubes as a liquid at 200C and leaves the tubes as saturated steam at 20 bar absolute. Calculate the rate (kg/h) at which steam is produced.
The rate at which steam is produced is equal to 701 kg/hour.
What is a Boiler?A Boiler may be characterized as a type of device or instrument that significantly transforms water into steam. There are two types of boiler are found. They are water tube boilers and fire tube boilers.
According to the question,
The power generated by combustion, W = 813kW.
The efficiency of the boiler, η = 65% = 0.65.
Temperature, To = 650°C.
Water enters the boiler tubes as a liquid, T1 = 20°C.
Water leaves the tubes as saturated steam, P2 = 20 bar.
The enthalpy of water at 20°C, [tex]h_1[/tex] = 83.9kJ/kg.
The enthalpy of water at 20 bar pressure, [tex]h_2[/tex] = 2797.29kJ/kg.
Enthalpy change can be calculated by ΔH = [tex]h_2-h_1[/tex]
= 2797.29kJ/kg - 83.9kJ/kg = 2713.3 kJ/kg.
The total energy that can be developed can be calculated by the formula:
Q = W × η = 813 × 0.65 = 528.45 kW.The mass of the flow rate of the rate at which steam is produced is calculated by the following formula:
[tex]m^.[/tex] = Q/ΔH= 528.45 kW/2713.3 kJ/kg.
= [tex]\frac{528.45kW }{2713.3kJ/kg} *\frac{3600kJ/h}{1kW}[/tex] = 701 kg/hour.
Therefore, the rate at which steam is produced is equal to 701 kg/hour.
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A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 5 in.2 and that approximately 2.3 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the time of submersion in years. The density of steel is 7.9 g/cm3.
Answer:
the estimated time of submersion is 17.7 years
Explanation:
Given the data in the question;
estimate the time of submersion in years.
we write down the relation between time of submersion and corrosion penetration as follows;
CPR(mpy) = K × W(mg) / [ A(in²) × p(g/cm³) × t(hr) ]
we solve for t
t = (K × W) / ( AP × CPR )
given that;
Area A = 5 in²
W = 2.3 kg = 2.3 × 10⁶ mg
density of steel p = 7.9 g/cm³
CPR = 200
we know that K is 534
so we substitute
t = (534 × 2.3 × 10⁶ mg) / ( 5 in² × 7.9 g/cm³ × 200 mpy )
t = 1,228,200,000 / 7900
t = 155468.3544 hr
t = 155468.3544 hr × ( 1 yrs / ( 365 × 24 hrs )
t = 17.7 years
Therefore, the estimated time of submersion is 17.7 years
A CUSTOMER BRINGS HER CAR INTO THE
SHOP WITH A COMPLAINT THAT BOTH
HEADLIGHTS ARE NOT WORKING.
TECHNICIAN A SAYS TO CHECK THE CONDITION OF THE BULBS FIRST. TECHNICIAN B CHECKS THE FUSE IN THE HEADLIGHT CIRCUIT & THEN TESTS THE OPERATION OF THE HEADLIGHT SWITCH WHO IS CORRECT AND WHY ?
HELP ME ASAP
a bc if the bulbs are in a bad conditio. than u know that u dont have to remove it but only repair it.
Consider steady flow of air through the device shown below (assume inviscid,
incompressible flow). The exit velocity is 100 ft/s and the differential pressure across the
nozzle upstream is 6 lb/ft?. Based on this information, determine (a) the height H of the water
in the manometer attached to the Pitot tube, and (b) the diameter d of the nozzle.
Answer:
im not for sure
Explanation:
The 150 mm thick wall of a gas fired furnace is constructed of fireclay brick (k=1.5 W/m.K) , tho=2600 kg/m3, and cp=1000 J/kg.K ) and is well insulated at its outer surface. The wall is at a uniform initial temperature of 20 degree C, when the burners are fired and inner surface is exposed to products of combustion for which T infinity=950 degree C and h=100 W/ m2.K.
(A) How long does it take for the outer surface of the wall to reach a temperature of 750 degree C?
(B) plot the temperature distribution in the wall at the foregoing time.
Answer:
I am thick but I dont know the anwser
The time that it will take for the outer surface of the wall to reach a temperature of 750 degree C will be 33800 seconds.
How to calculate the time?Using the approximation methods, Fo will be;
= In(0.215/1.262)/(1.4289)²
= 0.867
Then, the time taken will be:
= 0.867(0.15)²/(1.5/2600 × 1000)
= 33800 seconds.
In conclusion, the time taken is 33800 seconds.
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Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 ·K, deter- mine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at 45°C.
Answer:
Hello your question is incomplete attached below is the complete question
answer :
a) 95.80°C
b) 8.23 MW
Explanation:
Convection heat transfer coefficient = 860 W/m^2 . k
a) Calculate for the temp of sheet metal when it leaves the oil bath
first step : find the Biot number
Bi = hLc / K ------- ( 1 )
where : h = 860 W/m^2 , Lc = 0.0025 m , K = 60.5 W/m°C
Input values into equation 1 above
Bi = 0.036 which is < 1 ( hence lumped parameter analysis can be applied )
next : find the time constant
t ( time constant ) = h / P*Cp *Lc --------- ( 2 )
where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C
Input values into equation 2 above
t ( time constant ) = 0.10092 s^-1
Determine the elapsed time
T = L / V = 9/20 = 0.45 min
∴ temp of sheet metal when it leaves the oil bath
= (T(t) - 45 ) / (820 - 45) = e^-(0.10092 * 27 )
T∞ = 45°C
Ti = 820°C
hence : T(t) = 95.80°C
b) Calculate the required rate of heat removal form the oil
Q = mCp ( Ti - T(t) ) ------------ ( 3 )
m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s
Cp = 434 J/kg°C
Ti = 820°C
T(t) = 95.80°C
Input values into equation 3 above
Q = 8.23 MW