Answer:
Hello your question is incomplete below is the complete question
Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.
answer ; 311.36 k
Explanation:
Given data :
sum of mass * specific heat products for a base plate and components ( Mcp )
= 5000 J/K
effective heat transfer coefficient * surface area ( hA ) = 10 W/K
Initial temperature of plate and cooling air temperature( Tc ) = 295 k
power ( Q = W ) = 300 W
a) Determine plate temperature after 10 minutes
10 mins = 600 secs ( t )
heat supplied = change in temp + heat loss
Q * t = mCp ( ΔT ) + hA ( ΔT ) t
300*600 = 5000 * ( T -295 ) + 10 ( T -295 ) * 600
therefore ; T - 295 = 16.363
T = 311.36 K
Think about all the things that you have learned about wildfire in this module.
In light of climate change, explain how this risk map might look different in 100 years.
Answer:
CLIMATE CHANGE HAS inexorably stacked the deck in favor of bigger and more intense fires across the American West over the past few decades, science has incontrovertibly shown. Increasing heat, changing rain and snow patterns, shifts in plant communities, and other climate-related changes have vastly increased the likelihood that fires will start more often and burn more intensely and widely than they have in the past.
Explanation:
Laminated steel parts are designed:
A. to reduce noise and vibration.
B.for higher energy absorption.
C.to allow good part formability,
D. to increase tensile strength.
Answer:
to increase tensile strength.
Explanation:
Steel as part of the materials used in the construction industries, or other manufacturing industries could be defective if it was not appropriately manufactured. This defects could be due to foreign materials in it, scratches, blisters etc which would lead to the tensile strength being reduced.
In-order to overcome this, there is need to carry out a technique called lamination. This technique is the process of manufacturing the steel in multiple layers so that the composite materials improves the strength of the steel and its stability. This helps to prevent its failure when used in construction.
Technician A says that the B-pillars aid in resisting roof crush.
Technician B says that the IIHS rates vehicles to resist roof crush to five times the weight of the vehicle.
Who is right?
A only
O B only
Both A and B
Neither Anor B
Answer:
The answer is "Both A and B" are right
Explanation:
During the previous twenty years car producers have made significant advances in planning vehicle structures that give more noteworthy tenant insurance in planar accidents (Lund and Nolan 2003). Be that as it may, there has been little agreement with respect to the significance of rooftop strength in rollover crashes, just as the best strategy for surveying that strength. In 2006 one-fourth of lethally harmed traveler vehicle tenants were associated with crashes where vehicle rollover was considered the most hurtful occasion (Protection Establishment for Expressway Wellbeing, 2007). Numerous lethally harmed tenants in rollovers are unbelted, and some are totally or mostly launched out from the vehicle (Deutermann 2002).
There is difference concerning how underlying changes could influence launch hazard or the danger of injury for inhabitants who stay in the vehicle, paying little mind to belt use.
Answer:A Only
Explanation:
I Took the test
what is heat unit?in X ray machine
Answer:
joule
Explanation:
heat is express in joules in x Ray equipment
The reversible and adiabatic process of a substance in a compressor begins with enthalpy equal to 1,350 kJ/kg, and ends with enthalpy equal to 3,412 kJ/kg. If the compressor efficiency is 0.85, find the actual specific work required by the compressor to operate, in kJ/kg.
Answer:
the actual specific work required by the compressor to operate is 2425.88 kJ/kg
Explanation:
Given that;
h₁ = 1350 kJ/kg
h₂₅ = 3412 kJ/kg
compressor efficiency П_ise = 0.85
we know that;
compressor efficiency П_ise = isentropic work / actual work
П_ise = (h₂₅ - h₁) / (h₂ - h₁ )
so
0.85 = (h₂₅ - h₁) / (actual work )
Actual work = (h₂₅ - h₁) / 0.85
Actual work = (3412 - 1350) / 0.85
Actual work = 2062 / 0.85
Actual work = 2425.88 kJ/kg
Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to carry a minimum load of 13 N applied evenly to the top of the span. The maximum load for each member is 4.5 Newtons in compression and tension. Each member is 4.2 cm long (ball center to ball center). All joints must be fully constrained. Determine the max value of P before the truss fails. The section that can hold the most weight will receive bonus points.
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= [tex]\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12[/tex] = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
Scientists say that it takes over 50 years for trees to fully grow back after a fire. Most trees do not grow well under extreme environmental conditions.
If the climate continues to get warmer, do you think forests will be able to completely recover?
Yes
No
Explain your answer.
Answer:
no
Explanation:
How many trees would it take to reverse climate change?
1.2 trillion trees
Crowther says planting 1.2 trillion trees would give a reduction "way above" that figure. To put that in context, global CO2 emissions are around 37 billion tons per year.Apr 17, 2019
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot fro the seesaw to be balanced.
Given :
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.
To Find :
If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.
Solution :
We know, for seesaw to balance :
[tex]m_1gd_1=m_2gd_2[/tex]
Here, [tex]d_1\ and \ d_2[/tex] is distance from origin from pivot point.
Putting all the values in the equation, we get :
[tex]50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet[/tex]
Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.
Hence, this is the required solution.
Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribución de temperatura en función de "x" el cual es la distancia desde la cara izquierda de una pared, está dado por la siguiente función: T(x) = 150x2 -30x, donde x está en metros. Calcúlese el flujo de calor por unidad de área cuando x=0 y x=30, para cada caso menciones si se está enfriando o calentando el sólido.
Answer:
Para x=0:
[tex]\phi=1.2 W/m^{2}[/tex]
Para x=30 cm:
[tex]\phi=-2.4 W/m^{2}[/tex]
Explanation
Podemos utilizar la ley de Fourier par determinar el flujo de calor:
[tex]\phi=-k\frac{dT}{dx}[/tex](1)
Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.
Usando la ley de potencia para la derivda, tenemos:
[tex]\frac{dT(x)}{dx}=300x-30[/tex]
Remplezando esta derivada en (1):
[tex]\phi=-0.04(300x-30)[/tex]
Para x=0:
[tex]\phi=0.04(30)[/tex]
[tex]\phi=1.2 W/m^{2}[/tex]
Para x=30 cm:
[tex]\phi=-0.04(300*0.3-30)[/tex]
[tex]\phi=-2.4 W/m^{2}[/tex]
Espero que te haya ayudado!
A product (that has not been yet invented) that involves lighting up when something happens?
Answer:
Maybe when there is a fire there can be fire drones that can take it out. and it can also resuce people who are stuck there.
Explanation:
trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?
Answer:a
Ieieksdjd snsnsnsnsksks
Which of the following is a basic type of weld? O Groove O Lap O Edge O Corner
According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.
Answer:
high school diploma
Explanation:
According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.
In other words, they do not seek to acquire a post-secondary school education.
Answer:
B
Explanation:
According to edge its answer B
associate's degree or on-the-job experience
got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.
Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
bachelor’s degree
associate's degree or on-the-job experience .......This is the correct answer.
some college, no degree
associate degree
A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maximum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.
Answer:
W = 11,416.6879 N
L ≈ 64.417 cm
Explanation:
The maximum shear stress, [tex]\tau_{max}[/tex], is given by the following formula;
[tex]\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )[/tex]
[tex]t_w[/tex] = 1 cm = 0.01
h = 29 cm = 0.29 m
[tex]h_w[/tex] = 25 cm = 0.25 m
b = 15 cm = 0.15 m
[tex]I_c[/tex] = The centroidal moment of inertia
[tex]I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )[/tex]
[tex]I_c[/tex] = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴
Substituting the known values gives;
[tex]I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}[/tex]
[tex]I_c[/tex] = 1.2257083[tex]\bar 3[/tex] × 10⁻⁴ m⁴
From which we have;
[tex]4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )[/tex]
Which gives;
W = 11,416.6879 N
[tex]\sigma _{b.max} = \dfrac{M_c}{I_c}[/tex]
[tex]\sigma _{b.max}[/tex] = 1500 N/cm² = 15,000,000 N/m²
[tex]M_c[/tex] = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²
From Which we have;
[tex]M_{max} = \dfrac{W \cdot L}{4}[/tex]
[tex]L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417[/tex]
L ≈ 0.64417 m ≈ 64.417 cm.
How do you describe sound? (SELECT ALL THAT APPLY.) PLEASE HELP AND SELECT ALL THAT APPLY PLEASE!! A. Sound waves have to have a medium to travel through. B. The volume of a sound is known as amplitude. C. Loud sounds have high amplitude and vibrate with more energy than soft sounds. D. Sound waves are compression waves that cause energy transfer in air molecules.
Answer:
Sound waves are compression waves that cause energy transfer in air molecules
Sound waves have to have a medium to travel through
Loud sounds have high amplitude and vibrate with more energy than soft sounds
Explanation:
Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.
Sound is an example of a mechanical wave hence it requires a material medium for propagation.
The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.
calculate the quantities of materials required for the first class brickwork in 1:6 cement mortar for 10 cu.m. assume the suitable data.
advantage of iron meter
Answer:
Less Friction Error
Explanation:
The friction error is very less in the moving
iron instrument because their torque weight
ratio is high.
hope it helps!
Answer:
Universal Use -The MI instrument is independent of the of current and hence for both AC and DC
Explanation:
Mark as Brainlist Answer
Discuss why TVET Institutions need advice of the business community in order
to provide good programmes.
Answer:
Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community. To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town). This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace. Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.
Explanation:
TVET Institutions mean Technical and Vocational Education and Training Institutions. They play an important role in equipping young people to enter the world of work. They also continue to develop programs that will improve the employability of workers throughout their careers. They regularly respond to the changing labor market needs, adopt new training strategies and technologies, and expand the outreach of their training to current workers while grooming the young people for work.
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation technique is illustrated below where is the input, is the output and the ideal low-pass filter has a cutoff frequency of and a passband gain of 2. Determine . x(t) X(jω) = 0 g(t) = x(t)sin(2000πt) g(t) y(t) 2000π y(t) EENG 311: Signals and Systems Page 7 of 8 × lowpass filter y(t) cos(2000πt) g
Answer:
hello your question is incomplete attached below is the complete question
answer : attached below
Explanation:
let ; x(t) be a real value signal for x ( jw ) = 0 , |w| > 200[tex]\pi[/tex]
g(t) = x ( t ) sin ( 2000 [tex]\pi t )[/tex]
[tex]x_{1} (t) = \frac{1}{2} x(t) sin ( 4000\pi t )[/tex]
next we apply Fourier transform
attached below is the remaining part of the solution
Find R subscript C and R subscript B in the following circuit such that BJT would be in the active region with V subscript C E end subscript equals 5 V and I subscript C equals 25 m A V subscript C C end subscript equals 15 space V comma space V subscript D 0 end subscript equals 0.7 space V comma space beta equals 100 comma space V subscript A equals infinity.. Ignore the early effect in biasing calculations.
Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
A road is constructed at the capital cost of $6 million. At the end of Year 10, major improvements are to be made costing $17 million. At the end of Year 25, a replacement and upgrade is to be done at a cost of $29 million. At the end of year 40, the federal government issues a one-time tax credit in the amount of $12 million.
Over a 50-year analysis period (assuming a 10% interest rate) what is the annualized cost to the nearest dollar?
Answer:
The annualized cost is:
$299,272.
Explanation:
a) Data and Calculations:
Year 0 Capital cost of road construction = $6 million
Year 10 Major improvements cost = $17 million
Year 25 Replacement and upgrade cost = $29 million
Year 40 Federal government one-time tax credit = $12 million
Period of project analysis = 50 years
Cost of capital (discount rate) = 10%
Annualized cost at present value costs:
Amount spent Discount Factor Present value
$6 million 1 $6,000,000
$17 million 0.386 6,562,000
$29 million 0.092 2,668,000
($12 million) 0.0222 (266,400)
Total cost $14,963,600
Annualized cost = $14,963,600/50 = $299,272
b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50. Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%. And then, the average of the cost is obtained by dividing the total net present cost into 50 years.
A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
Answer:
a) for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b) for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Explanation:
Given that;
A₂ = 0.001 m²
P₁ = 1 MPa
T₁ = 360 K
k = 1.4
P₂ = 500 Kpa
(1000/500)^(1.4-1 / 1.4) = 360 /T₂
2^(0.4/1.4) = 360/T₂
1.219 = 360 / T₂
T₂ = 360 / 1.219
T₂ = 295.32 K
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
we substitute
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
1.005 × 360 = 1.005 × 295.32 + v₂²/2000
v₂ = 360.56 m/s²
p₂v₂ = mRT₂
500 × (0.001 × 360.56) = m × 0.287 × 295.32
m = 2.127 kg/s
so Mach Number = V₂ / Vc
Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s
So Mach Number = V₂ / Vc = 360.56 / 344.47 = 1.046
Therefore for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b)
AT P₂ = 784 kPa
(1000/784)^(1.4-1 / 1.4) = 360/T₂
T₂ = 335.82 K
now
V₂²/2000 = 1.005( 360 - 335.82)
V₂ = 220.45 m/s
P₂V₂ = mRT₂
784 × (0.001 × 220.45) = m( 0.287) ( 335.82)
172.83 = 96.38 m
m = 172.83 / 96.38
m = 1.793 kg/s
just like in a)
Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s
Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6
Therefore for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Following are the
Given:
[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]
To find:
Flow rate of mass, and Mach number
Solution:
For point a)
Using formula:
[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]
[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]
Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]
[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]
[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]
For point b)
[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]
now
[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]
Learn more about flow rate of mass, and Mach number:
brainly.com/question/15055873
I need help with part (C). Pleasee help me. It’s due in a few hours.
Answer:
u do the same thing as part B but only add 100 k, I think, cuz I'm still in middle school but I mean if u see it asks u to do the same thing as B but C says that instead, u do it at half pressure and 100 k is higher temp so what its asking is to repeat b but the twist is u do it at half pressure and 100 k is the higher temp
hope this helps :)
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.
What is forging?Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.
Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.
The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.
Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.
Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.
Thus, the answer is cold forging.
For more details regarding forging, visit:
https://brainly.com/question/14235193
#SPJ2
using credit reduces future income
Answer:
lol
Explanation:r
How many kg moles of Sodium Sulphate will contain 10 kg of
Sodium?
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
(as per my thinking)
Answer:
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
Consider a condenser in which steam at a specified temperature is condensed by rejecting heat to the cooling water. If the heat transfer rate in the condenser and the temperature rise of the cooling water is known, explain how the rate of condensation of the steam and the mass flow rate of the cooling water can be determined. Also, explain how the total thermal resistance R of this condenser can be evaluated in this case.
Answer:
Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]
(ΔT)[tex]_{cooling water}[/tex] and Q is given
[tex]m_{cooling water}[/tex] = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]
next the rate of condensation of the steam
Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]
[tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]
Total resistance of the condenser is
R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]
Explanation:
How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined
Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]
(ΔT)[tex]_{cooling water}[/tex] and Q is given
[tex]m_{cooling water}[/tex] = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]
next the rate of condensation of the steam
Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]
[tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]
Total resistance of the condenser is
R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]
A battery with an f.e.m. of 12 V and negligible internal resistance is connected to a resistor of 545 How much energy is dissipated by the resistor in 65 s?
Answer:
When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then
R
1
in Figure 1(a) could be the resistance of the screwdriver’s shaft,
R
2
the resistance of its handle,
R
3
the person’s body resistance, and
R
4
the resistance of her shoes.
Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)
Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.
Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).
To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.
According to Ohm’s law, the voltage drop,
V
, across a resistor when a current flows through it is calculated using the equation
V
=
I
R
, where
I
equals the current in amps (A) and
R
is the resistance in ohms
(
Ω
)
. Another way to think of this is that
V
is the voltage necessary to make a current
I
flow through a resistance
R
.
So the voltage drop across
R
1
is
V
1
=
I
R
1
, that across
R
2
is
V
2
=
I
R
2
, and that across
R
3
is
V
3
=
I
R
3
. The sum of these voltages equals the voltage output of the source; that is,
V
=
V
1
+
V
2
+
V
3
.
This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation
P
E
=
q
V
, where
q
is the electric charge and
V
is the voltage. Thus the energy supplied by the source is
q
V
, while that dissipated by the resistors is
q
V
1
+
q
V
2
+
q
V
3
.
Explanation:
What type of engineer makes sure equipment is safe and operational
Answer:
mechanical engineer is the best answer
Technician A says that an A-pillar may be designed to transfer collision energy
Technician B says that a floor pan reinforcement may be designed to transfer collision energy
Who is right?
A. Aonly
B. B only
C. Both A and B
D. Neither Anor B
Technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.
What is Collision energy?Collision energy may be defined as a circumstance in which two or more bodies or particles come together with a resulting exchange of energy and alteration of direction.
Both pillar and floor pan reinforcement may be designed to transfer collision energy through the most prominent way to explain this mechanism of collisions.
A pillar exerts complete pressure on the floor in order to support the roof of the building, while floor pan reinforcement performs the same mechanism of collision energy with a different mechanism of action.
Therefore, technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.
To learn more about Collisions energy, refer to the link:
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