Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100), ( b ) (110), and ( c ) (111) planes.
Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d = [tex]\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}[/tex]
a)(100)
a=5.28 Å
Distance = [tex]\frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}[/tex]=5.28 Å
b)(110)
Distance = [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}}[/tex] = 3.73 Å
c)(111)
Distance= [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}[/tex]= 3.048 Å
In each case indicate whether the quantity in question increased, decreased or stayed the same when the string length is increased. Assume that the tension is unchanged. The function generator is kept at the same frequency, and the string is in resonance in all cases. Part A Number of antinodes ___ Number of antinodes ___ increased. decreased. stayed the same. Request Answer Part B Wavelength ___ Wavelength ___ increased. decreased. stayed the same. Request Answer Part C Fundamental frequency ___ Fundamental frequency ___ increased. decreased. stayed the same.
Answer:
Answer: No of anti-nodes increases
Answer: wavelength remains same.
Answer: fundamental frequency decreases
Explanation:
a)
The number of nodes (n) would have (n-1) anti-nodes.
The relation of Length of string with n is given below:
[tex]L = \frac{n*lambda}{2}[/tex]
Hence, n and L are directly proportional so as string length increases number of nodes and anti-nodes also increases.
Answer: No of anti-nodes increases
b)
Wavelength is dependent on the frequency:
[tex]lambda = \frac{v}{f}[/tex]
The speed v of the string remains same through-out and frequency of generator is unchanged!
Hence according to above relationship lambda is unchanged.
Answer: wavelength remains same.
C
Fundamental frequency equates to 1st harmonic that 2 nodes and 1 anti-node. Wavelength is = 2*L
Hence, if L increases wavelength increases
Using relation in part b
As wavelength increases fundamental frequency decreases
Answer: fundamental frequency decreases
Consider a point in a structural member that is subjected to plane stress. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are Sx = 175 MPa, Sy = 90 MPa, and Sxy = 75 MPa.
Answer:
C = 132.5 MPa
R = 86.20 MPa
Explanation:
Given
σx = 175 MPa
σy = 90 MPa
τxy = 75 MPa
For the given state of stress at a point in a structural member, determine the center C and the radius R of Mohr’s circle.
We apply the following equation for the center C
C = (σx + σy) / 2
C = (175 MPa + 90 MPa) / 2
C = 132.5 MPa
The Radius can be obtained as follows
R = √(((σx - σy) / 2)² + (τxy)²)
R = √(((175 MPa - 90 MPa) / 2)² + (75 MPa)²)
R = 86.20 MPa
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 13.2 ft/sft/s at point A and 17.6 ft/sft/s at point C. The cart takes 3.00 ss to go from point A to point C, and the cart takes 1.90 ss to go from point B to point C. What is the car's speed at point B?
Answer:
15.99 ft/s
Explanation:
From Newton's equation of motion, we have
v = u + at
v = Final speed
u = initial speed
a = acceleration
t = time
now
for the points A and C
v = 17.6 ft/s
u = 13.2 ft/s
t = 3 s
thus,
17.6 = 13.2 + a(3)
or
3a = 17.6 - 13.2
3a = 4.4
or
a = 1.467 m/s²
Thus,
For Points A and B
v = speed at B i.e v'
u = 13.2 ft/s
a = 1.467 ft/s²
t = 1.90 s
therefore,
v' = 13.2 + (1.467 × 1.90 )
v' = 13.2 + 2.7867
v' = 15.9867 ≈ 15.99 ft/s
Check my work Check My Work button is now disabledItem 16Item 16 3 points As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.
Answer:
W = 1.7593 * 10 ^ (-7) KJ
Explanation:
The work done by the bubble is given:
[tex]W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ[/tex]
Answer: W = 1.7593 * 10 ^ (-7) KJ
Design a sequential circuit with two D flip-flops A and B, and one input x_in.
(a)When xin = 0, the state of the circuit remains the same. When x_in = 1, the circuit goes through the state transitions from 00 to 01, to 11, to 10, back to 00, and repeats.
(b)When xin = 0, the state of the circuit remains the same. When x_in =1, the circuit goes through the state transitions from 00 to 11, to 01, to 10, back to 00, and repeats.
Answer:
View Image
Explanation:
The question is basically asking you to build a 2-bit asynchronous counter.
What the counter does is it increase it's value by 01₂ every clock pulse. So at 0₂, nothing happens, but at 1₂ it'll count up by 1. It then reset to 00₂ when it overflows.
The design for it is pretty much universal so I kinda did this from memory.
a.) A count-up counter (from 00-11) is simply made by connecting Q' to D, and the output of the previous DFF to the clock of the next one.
b.) A count-down counter (from 11-00) is simply made by using the same circuit as the count-up counter, but you connect Q' to the clock instead of Q.
A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.
a. What is the energyon in terms of k adm
b. If the oscillator is a bonded atom with k = 15 N/m and m = 4 × 10-26 kg, what is the frequency (Hz) of the emitted photon? (Note: the energy of a photon is Ephoton= hf)
c. In which region of the electromagnetic spectrum (x-ray, visible, microwave, etc.) does this photon belong?
The answer explains the energy of a quantum harmonic oscillator, calculates the frequency of an emitted photon, and identifies the region of the electromagnetic spectrum it belongs to.
Explanation:a. Energy: The energy of a quantum harmonic oscillator can be represented as En = (n+1/2)h(sqrt(k/M)), where n = 0,1,2... and h represents Planck's constant.
b. Frequency Calculation: Using the given values of k = 15 N/m and m = 4 x 10^-26 kg, you can calculate the frequency of the emitted photon using the formula w = sqrt(k/M)/(2pi).
c. Electromagnetic Spectrum: To determine the region of the electromagnetic spectrum the photon belongs to, compare the frequency calculated to the known ranges of various regions like x-ray, visible, and microwave.
a 100mh inductor is placed parallel with a 100 ohm resistor in the circuit , the circuit has a source voltage of 30 vac and a frequancy of 200 Hz what is the current through the inductor
Answer:
current through the inductor = 0.24 A
Explanation:
given data
inductor = 100mh
resistor = 100 ohm
voltage = 30 vac
frequancy = 200 Hz
to find out
current through the inductor
solution
current through the inductor will be when inductor is place parallel with resistor
across resistor
XL = i2πl
XL = i2π×200×100×[tex]10^{-3}[/tex] = i126.66
so
current through the inductor = [tex]\frac{voltage}{XL}[/tex]
current through the inductor = [tex]\frac{30}{125.66}[/tex]
current through the inductor = 0.24 A
Assume a strain gage is bonded to the cylinder wall surface in the direction of the hoop strain. The strain gage has nominal resistance R0 and a Gage Factor GF. It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage, Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.)Calculate the voltage change ∆V across the Wheatstone bridge when the cylinder is pressurized to ∆P = 2.5 atm. Assume the vessel is made of 3004 aluminum with height h = 21 cm, diameter d = 9 cm, and thickness t = 65 µm. The Gage Factor is GF = 2 and the Wheatstone bridge has Vin = 6 V. The strain gage has nominal resistance R0 = 120 Ω.
Answer:
5.994 V
Explanation:
The pressure as a function of hoop strain is given:
[tex]P = \frac{4*E*t}{D}*\frac{e_{h} }{2-v}[/tex]
[tex]e_{h} = \frac{D*P*(2-v)}{4*E*t} .... Eq1[/tex]
For wheat-stone bridge with equal nominal resistance of resistors:
[tex]V_{out} = \frac{GF*e*V_{in} }{4} .... Eq2[/tex]
Hence, input Eq1 into Eq2
[tex]V_{out} = \frac{GF*e*V_{in}*D*P*(2-v) }{16*E*t} .....Eq3\\[/tex]
Given data:
P = 253313 Pa
D = d + 2t = 0.09013 m
t = 65 um
GF = 2
E = 75 GPa
v = 0.33
Use the data above and compute Vout using Eq3
[tex]V_{out} = \frac{2*6*0.09013*253313*(2-0.33) }{16*75*10^9*65*10^-6} \\\\V_{out} = 0.006285 V\\\\change in V = 6 - 0.006285 = 5.994 V[/tex]
If a barrel of oil weighs 1.5 kN, calculate the specific weight, density, and specific gravity of the oil. The barrel weighs 110 N
Answer
given,
oil barrel weight = 1.5 k N = 1500 N
weight of the barrel = 110 N
Assuming volume of barrel = 0.159 m³
weight of oil = 1500-110
= 1390 N
[tex]specific\ weight = \dfrac{weight}{volume}[/tex]
[tex]specific\ weight = \dfrac{1390}{0.159}[/tex]
= 8742.14 N/m³
[tex]mass = \dfrac{weight}{g}[/tex]
[tex]mass = \dfrac{1390}{9.8}[/tex]
= 141.84 kg
[tex]density = \dfrac{mass}{volume}[/tex]
[tex]density = \dfrac{141.84}{0.159}[/tex]
= 892.05 kg/m³
[tex]Specific\ gravity = \dfrac{density\ of\ oil}{density\ of\ water}[/tex]
[tex]Specific\ gravity = \dfrac{892.05}{1000}[/tex]
= 0.892
When generating a hierarchical cluster, a tree structure called a _______.
Answer:
Dendrogram
Explanation:
Dendrogram is referred to as the tree structure that represents the hierarchy between the object in a cluster. it is also referred to as an output that is drawn from clustering.
The dendrogram is interpreted by observing the object situated at the higher side in a scatter plot. By joining the object at the same level it represents the order of cluster. another purpose of modifying form of Dendrogram is to help in calculating the distance between the objects in clusters.
Write IEEE floating point representation of the following decimal number. Show your work.
1.25
Answer:
00111111101000000000000000000000
Explanation:
View Image
0 01111111 01000000000000000000000
The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.
The next 8-bits are for the exponents.
The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.
And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.
You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰
The exponent bit value is 127+0=127 which is 01111111 in binary.
The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:
010...00000000000000000000
A room is cooled by circulating chilled water through a heat exchanger located in the room. The air is circulated through the heat exchanger by a 0.25-hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25-hp equipment is 60 percent. Determine the rate of heat supply by the fan–motor assembly to the room.
To determine the rate of heat supply by the fan-motor assembly, the electrical input power is calculated based on the 0.25-hp shaft output and 60% efficiency of the motor. The resulting heat supply to the room is the same as the electrical power input, which is 310.83 watts.
Explanation:The question asks to determine the rate of heat supply by a fan-motor assembly used to circulate chilled water through a heat exchanger for cooling a room. Given that the fan has a shaft output of 0.25 horsepower (hp) and that small electric motors driving such equipment typically have an efficiency of 60 percent, we can calculate the electrical power input needed to run the fan.
The electrical power input (Pinput) can be calculated as:
Pinput = Poutput / Efficiency
Where Poutput is the shaft output power (0.25 hp) and 'Efficiency' is the efficiency of the electric motor (60%, or 0.60 in decimal form).
Pinput = (0.25 hp) / 0.60
To convert horsepower to watts, we use the conversion factor 1 hp = 746 watts.
Pinput = (0.25 hp × 746 watts/hp) / 0.60
Pinput = 310.83 watts (rounded to two decimal places)
The rate of heat supply to the room will be equal to the electrical power input to the fan, which is 310.83 watts. This accounts for both the useful work done and all inefficiencies in the system that convert electrical energy into heat.
In a residential heat pump, Refrigerant-134a enters the condenser at 800 kPa and 50oC at a rate of 0.022 kg/s and leaves at 750 kPa subcooled by 3oC. The refrigerant enters the compressor at 200 kPa superheated by 4oC. Determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat pump. Also, (d) determine the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 200 kPa and 800 kPa.
Answer:
a) 0.813
b) 4.38 KW
c) COP = 5.13
d) 3.91 KW , COP = 6.17
Explanation:
Data obtained A-13 tables for R-134a:
[tex]h_{1} = 247.88 \frac{KJ}{kg} \\s_{1} = 0.9575 \frac{KJ}{kgK}\\h_{2s} = 279.45 \frac{KJ}{kg}\\h_{2} = 286.71 \frac{KJ}{kg}\\h_{3} = 87.83 \frac{KJ}{kg}[/tex]
The isentropic efficiency of the compressor is determined by :
[tex]n_{C} = \frac{h_{2s} - h_{1} }{h_{2} - h_{1} }\\= \frac{279.45 - 247.88 }{286.71 - 247.88}\\= 0.813[/tex]
The rate of heat supplied to the room is determined by heat balance:
[tex]Q = m(flow) * (h_{2} -h_{3})\\= (0.022)*(286.71 - 87.83)\\\\= 4.38KW[/tex]
Calculating COP
[tex]COP = \frac{Q_{H} }{W} \\COP = \frac{Q_{H} }{m(flow) * (h_{2} - h_{1}) }\\\\COP = \frac{4.38}{(0.022)*(286.71-246.88)}\\\\COP = 5.13[/tex]
Part D
Data Obtained:
[tex]h_{1} = 244.5 \frac{KJ}{kg} \\s_{1} = 0.93788 \frac{KJ}{kgK}\\h_{2} = 273.31 \frac{KJ}{kg}\\h_{3} = 95.48 \frac{KJ}{kg}[/tex]
The rate of heat supplied to the room is determined by heat balance:
[tex]Q = m(flow) * (h_{2} -h_{3})\\= (0.022)*(273.31 - 95.48)\\\\= 3.91KW[/tex]
Calculating COP
[tex]COP = \frac{Q_{H} }{W} \\COP = \frac{Q_{H} }{m(flow) * (h_{2} - h_{1}) }\\\\COP = \frac{4.38}{(0.022)*(273.31-244.5)}\\\\COP = 6.17[/tex]
The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.
a. What is the forward thrust of this engine?
b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?
c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?
Answer:
T = 5416.67 N
T = -2083.5 N
T = 0
Explanation:
Forward thrust has positive values and reverse thrust has negative values.
part a
Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N[/tex]
Answer: The thrust force T = 5416.67 N
part b
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:
[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N[/tex]
Answer: The thrust force T = -2083.5 N reverse direction
part c
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.
Answer: T = 0 N
My Notes How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid?
How many grams of water are in the same solution?
Answer:
a)29.9 b) 9.81
Explanation:
Wt% = mass of solute / mass of solvent × 100
0.749 = mass of solute / mass of solvent
a) Mass of perchloric acid = 0.749 × 39.1 = 29.29
b) Mass of water = 39.1 - 29.29 = 9.81
Electrical failure can lead to ________failure, which in turn can lead to software failure.
Answer:
Electrical failure can lead to _hardware__failure, which in turn can lead to software failure
One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight last?
Answer:
time = 5.22 hr
Explanation:
Given data:
Energy of battery = 9400 J
Power consumed by three led bulb is 0.5 watt
we know Power is give as
[tex]Power = \frac{ energy}{time}[/tex]
plugging all value and solve for time
[tex]time = \frac{energy}{power}[/tex]
[tex]time = \frac{9400}{0.5}[/tex]
time = 18,800 sec
in hour
1 hour = 3600 sec
therefore in 18,800 sec
[tex]time = \frac{18800}{3600} = 5.22 hr[/tex]
Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at top equal to 0.5 the area of the tension steel at the bottom.
Explanation:
Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete
The net potential energy between two adjacent ions, EN, may be represented by EN = -A/r + B/rn Where A, B, and n are constants whose values depend on the particular ionic system. Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure: 1. Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of EN versus r is a minimum at E0. 2. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing. 3. Determine the expression for E0 by substitution of r0 into the above equation for EN. What is the equation that represents the expression for E0?
The equation that represents the expression for the bonding energy [tex](E_0\))[/tex] in terms of the parameters A, B, and n is: [tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex].
Step 1: Differentiate EN with Respect to r
Given:
[tex]\[ E_N = -\frac{A}{r} + \frac{B}{r^n} \][/tex]
Differentiate [tex]\(E_N\)[/tex] with respect to r:
[tex]\[ \frac{dE_N}{dr} = \frac{A}{r^2} - \frac{nB}{r^{n+1}} \][/tex]
Set the derivative equal to zero to find the minimum (equilibrium) point:
[tex]\[ \frac{A}{r^2} - \frac{nB}{r^{n+1}} = 0 \][/tex]
Solve for r as
[tex]\[ \frac{A}{r^2} = \frac{nB}{r^{n+1}} \][/tex]
[tex]\[ r^n = \frac{nB}{A} \][/tex]
[tex]\[ r = \left(\frac{nB}{A}\right)^{\frac{1}{n}} \][/tex]
Step 2: Solve for r in terms of A, B, and n
The equilibrium interionic spacing [tex]\(r_0\)[/tex] is the value of r at the minimum point,
[tex]\[ r_0 = \left(\frac{nB}{A}\right)^{\frac{1}{n}} \][/tex]
Step 3: Determine the Expression for E0
Substitute [tex]\(r_0\)[/tex] into the expression for [tex]\(E_N\):[/tex]
[tex]\[ E_0 = -\frac{A}{r_0} + \frac{B}{r_0^n} \][/tex]
[tex]\[ E_0 = -\frac{A}{\left(\frac{nB}{A}\right)^{\frac{1}{n}}} + \frac{B}{\left(\frac{nB}{A}\right)^{\frac{n}{n}}} \][/tex]
Simplify:
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{B}{(nB)^{\frac{n}{n}}} \][/tex]
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{B}{nB} \][/tex]
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex]
Combine the terms:
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex]
So, the equation is [tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex].
Learn more about Equilibrium here:
https://brainly.com/question/30694482
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A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at 2 bar to a final volume of 0.12 m^3, while being slowly heated through the base. The change in internal energy of the gas is 0.25 kJ. the piston and cylinder walls are fabricated from heat-resistant material and the piston moves smoothly in the cylinder. The local atmosphere pressure is 1 bar.
a.) For the gas as the system, evaluate the work and the heat transfer, each in
b.) For the piston as a syatem, evautate the work and change in potential energy, in kJ.
Answer: (a). W = 4KJ and Q = 4.25KJ
(b). W = -2KJ and ΔPE = 2KJ
Explanation:
(a).
i. We are asked to calculate the work done during the expansion process considering gas as system.
from W = [tex]\int\limits^a_b {p} \, dV[/tex] where a = V₂ and b = V₁
so W = P(V₂-V₁)
W = (2 × 10²) (0.12 - 0.10)
W = 4 KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0
∴ Q - W = ΔU
Q = ΔU + W
Q = 0.25 + 4
Q = 4.25 KJ
(b).
i. to calculate the work done during the expansion process considering piston as system;
W = [tex]\int\limits^a_b {(Patm - Pgas)} \, dV[/tex]where a and b represent V₂ and V₁ respectively.
W = (Patm - Pgas)(V₂ - V₁)
W = (1-2) ×10² × (0.12-0.1)
W = -2KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
Q = 0 since the piston and cylinder walls are perfectly insulated.
for piston, we neglect the change in internal energy and kinetic energy
ΔU = ΔKE = 0
from Q - W = ΔU + ΔKE + ΔPE
0 - (-2) = 0 + 0 + ΔPE
∴ ΔPE = 2KJ
A) Fix any errors to get the following program to run in your environment. B) Document each line of code with comments and describe any changes you had to make to the original code to get it to work. C) Write a summary of what your final version of the program does.
You may also add white space or reorder the code to suit your own style as long as the intended function does not change.
#include
using namespace std;
void m(int, int []);
void p(const int list[], int arraySize)
{
list[0] = 100;
}
int main()
{
int x = 1;
int y[10];
y[0] = 1;
m(x, y);
cout << "x is " << x << endl;
cout << "y[0] is " << y[0] << endl;
return 0;
}
void m(int number, int numbers[])
{
number = 1001;
numbers[0] = 5555;
}
Answer:
Answer is explained below
Explanation:
Part A -:
Error statement -:
/*
prog.cpp: In function ‘void p(const int*, int)’:
prog.cpp:7:15: error: assignment of read-only location ‘* list’
list[0] = 100;
*/
There is one error in the code in following part
void p( const int list[], int arraySize)
{
list[0] = 100;
}
you are passing list as constant but changing it inside the function that is not allowed. if you pass any argument as const then you can't change it inside the function. so remove const from function argument solve the error.
Part B -:
change made
void p( int list[], int arraySize)
{
list[0] = 100;
}
Executable fully commented code -:
#include <iostream> // importing the iostream library
using namespace std;
void m(int, int []); // Function declearation of m
void p( int list[], int arraySize) // definition of Function p
{
list[0] = 100; // making value of first element of list as 100
}
int main()
{
int x = 1; // initilizing x with 1
int y[10]; // y is a array of 10 elements
y[0] = 1; // first element of y array is 1
m(x, y); // call m function
// printing the desired result
cout << "x is " << x << endl;
cout << "y[0] is " << y[0] << endl;
return 0;
}
void m(int number, int numbers[]) // Function definition of m
{
number = 1001; // value of number is 1001 locally
numbers[0] = 5555; // making value of first element of numbers array 5555
}
Part C :-
In program we initilize x with value 1 and create an array y of 10 elements.
we initilize the y[0] with 1\
then we call function m. In function m ,first argument is value of x and second argument is the pointer to the first element of array y.
so value of x is changed locally in function m and change is not reflected in main function.
but value of y[0] is changed to 5555 because of pass by refrence.
So we are getting the following result :-
x is 1
y[0] is 5555
The gage pressure in a liquid at a depth of 3 m is read to be 48 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.
The gage pressure in the same liquid at a depth of 9 m is__________ kPa.
Answer: 144kpa
Explanation:
pressure density =p
The Gage pressure (P1)=48kpa
The height (h1)=3m
The Gage pressure (P2)=?
The height (h2)=9m
P=p * g * h
P1/P2=p * g * h1/p * g * h2
Cancelling out similar terms:
Therefore, P1/P2=h1/h2
P2=P1*h2/h1
Hence, P2=48*9/3=144kpa.
Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) Cylindrical wall (b) Plane shell (c) Spherical shell (d) All of them (e) None of them
Answer:
Plane shell which is option b
Explanation:
The temperature in the case of a steady one-dimensional heat conduction through a plane wall is always a linear function of the thickness. for steady state dT/dt = 0
in such case, the temperature gradient dT/dx, the thermal conductivity are all linear function of x.
For a plane wall, the inner temperature is always less than the outside temperature.
A glass window pane, 1 m wide, 1.5 m high, and 5 mm thick, has a thermal conductivity of kg = 1.4 W/(m∙K). On a cold winter day, the indoor and outdoor temperatures are 15 °C and −25 °C respectively. (a) For a single-pane window at steady state, what is the rate of heat transfer through the glass? (10 pts) (b) To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by a dead-air space. The thermal conductivity of air is ka = 0.024 W/(m∙K). If the spacing between the two glasses is 10 mm. Calculate the temperatures of the glass surfaces in contact with the dead-air at the steady state. (10 pts) (c) Calculate the heat loss through the double-pane window in (b). (5 pts)
Answer:
a) Rate of heat transfer = 16.8KW
b) Temperature of glass surface = 15degree Celsius
c) Heat loss through frame = 141.34W
Explanation:
The concept used to approach this question is the Fourier's law of head conduction postulated by Joseph Fourier. it states that the rate of heat flow through a single homogeneous solid is directly proportional to the area and to the direction o heat flow and to the change in temperature with respect to the path length. Mathematically,
Q = -KA dt/dx
The detailed and step by step calculation is attached below
Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 300°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 3560 kJ and kinetic and potential energy effects are negligible.
Determine the final volume, in m3, and the work for the process, in kJ.
To determine the final volume and work done during the heating of water in a piston-cylinder assembly at constant pressure, tabulated data like steam tables are required since water is not an ideal gas. The work done is calculated using the formula W = PΔV.
Explanation:The student has been asked to find the final volume and the work done during a constant pressure process in which 5 kg of water is heated in a piston-cylinder assembly from an initial state of 5 bar and 300°C. To solve for the final volume and work done, one would typically use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat input minus the work output. However, since water at this state is not an ideal gas, tabulated data from steam tables or software would be used to determine the specific volume at the final state and then multiplied by the mass to find the total volume. The work done in a constant pressure process is equal to the pressure times the change in volume (W = PΔV). Without the final specific volume from the tables, we cannot compute the final volume or work directly.
(1 point) Consider the initial value problem 2????y′=4y, y(−1)=−2. Find the value of the constant ???? and the exponent ???? so that y=???????????? is the solution of this initial value problem. y= help (formulas) Determine the largest interval of the form ????<????<???? on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution. help (inequalities) What is the actual interval of existence for the solution (from part a)? help (inequalities)
Answer:
Here is the missing part of the question ;
find the value of the constant C and the exponent r so that y = Ctr is the solution of this initial value problem. Determine the largest interval of the form a<t<b
Explanation:
The step by step explanation is as given in the attachment. You will notice I used β as the exponential in place of r.What is the standard half-cell potential for the oxidation of methane under acidic conditions? The reaction for methane is as follows:
CH4(g) + 2H20(l) → CO2 +8H⁺ +8e⁻
What element is oxidized and how does its oxidation state change?
Answer:
The element that is oxidized is carbon.
Its oxidation state increased. It increased from -4 to +4
Explanation:
Oxidation is a process that involves increase in oxidation number.
The oxidation number of carbon in CH4 is -4
C + (1×4) = 0
C + 4 = 0
C = 0 - 4 = -4
The oxidation number of carbon in CO2 is +4
C + (2×-2) = 0
C - 4 = 0
C = 0+4 = 4
Increase in the oxidation number of carbon from -4 to +4 means carbon is oxidized
The method of breaking digital messages into small fixed bundles, transmitting them along different communication paths, and reassembling them at their destination is called:
Answer:
Packet switching
Explanation:
Packet switching - it is referred to as the transmission process which involved sending of packets through communication path and last reassembling them again.
The sending of packets is done by broken the digital message into required pieces for efficient transfer.
in this process, every individual parcel is sent separately.
If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the conductor.
Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
Surface Charge Density = 26.55 x 10^-12 C/m²
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
The surface charge density on the conductor is; σ = 13.275 × 10⁻¹² C/m²
What is the surface charge density?The formula for surface charge density on a conductor in an electric field just outside the surface of conductor is;
σ = E * ϵ₀
where;
E is electric field = 1.5 N/C
ϵ₀ is permittivity of space = 8.85 × 10⁻¹² C²/N.m²
Thus;
σ = 1.5 * 8.85 × 10⁻¹²
σ = 13.275 × 10⁻¹² C/m²
Read more about Surface Charge Density at; https://brainly.com/question/14306160