A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. Technician A is correct while Technician B is wrong.
What is a voltmeter?A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. It is linked in parallel. It generally has a high resistance and draws very little current from the circuit.
Given the technician connects the voltmeter in parallel across two points in a circuit. Therefore, the technician A is correct the meter will provide a reading of the potential difference in volts.
Hence, Technician A is correct while Technician B is wrong.
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Look back at the Time Traveller's dialogue. Explain what it reveals about him and about what has happened to him. In your response, include at least one of the phrases he uses.
Answer:
The time traveler feels special for having performed a feat most men will never be able to. In his statement "I’ve lived eight days . . . such days as no human" he clearly feels very unique and at such demands to be respected and given special previledges which is seen when he asks ‘Where’s my mutton?’ upon been offered a special position to sit, declaring his right.
The time traveler is seen to capitalize and make gain off his feat such that he demands for meat in exchange for his story and also, he changed his stance on eating first before telling his story when he was offered money for his story as he ate; reverting to apologising by saying ‘I suppose I must apologize' and proceeded with telling the story as he ate.
In the dialogue, The time traveler is said to feel special and unique. He wanted all to respected and given him preferential treatment such as when he asks ‘Where’s my mutton?’.
In the story, the traveler is one that often uses different means to make profits for himself.Who is a time Traveler?This is known to be a person that is said to have went back in time or between different times in the past or maybe in the future.
This kind of people are often strange and do engage in unnatural things and in this story, the time traveler sees himself as is someone who is special and unique.
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Water flows with a velocity of 3 m/s in a rectangular channel 3 m wide at a depth of 3 m. What is the change in depth and in water surface elevation produced when a gradual contraction in the channel to a width of 2.6 m takes place? Determine the greatest contraction allowable without altering the specified upstream conditions.
Answer: new depth will be 3.462m and the water elevation will be 0.462m.
The maximum contraction will be achieved in width 0<w<3
Explanation:detailed calculation and explanation is shown in the image below
A 300 mm long steel bar with a square cross section (25 mm per edge) is pulled in tension with a load of 84998 N , and experiences an axial elongation of 0.18 mm . Assuming that the deformation is entirely elastic, calculate the elastic modulus of this steel in GPa.Answer Format: X (no decimal places)
Answer:
Elastic Modulus = 227 GPa
Explanation:
Given,
Load = 84998 N
Length of bar = 300 mm = 0.3 m
Elongation = 0.18 mm = 0.00018 m
Cross sectional Area of the bar = (25mm × 25mm) = 0.025 × 0.025 = 0.000625 m²
From Hooke's law, the stress experienced by a material is proportional to the strain experienced by the same body, as long as the elastic limit isn't exceeded.
Stress ∝ strain
The coefficient of proportionality is the elastic modulus, E.
Stress = E × (Strain)
Stress = (Load)/(Cross sectional Area)
Stress = (84998 ÷ 0.000625) = 135,996,800 N/m²
Strain = (Change in length)/(Original length)
Strain = (ΔL/L) = 0.00018 ÷ 0.3 = 0.0006
E = (Stress/Strain)
E = 135,996,800 ÷ 0.0006 = 226,661,333,333.3 Pa = (2.267 × 10¹¹) Pa
1 GPa = 10⁹ Pa
(2.267 × 10¹¹) = 2.267 × 10² × 10⁹ = 226.7 GPa = 227 GPa to the nearest GPa. (No decimal place)
Hope this Helps!!!
Answer:
227 Gpa
Explanation:
∆L = PL/AE
E = PL/A∆L
E is Elastic Modulus
L is length
A is Area
L = 300 mm = 300* 10^-3
A = (25 * 10 ^-3)^2
P = 84998N
∆L = 0.18mm = 0.18*10^-3
E = 84998*300*10^-3/((25*10^-3)^2*0.18*10^-3
E = 226661333333.3Pa
= 226.7 * 10^9Pa
10^9Pa = 1 GPA
E = 226.7 Gpa
E = 227 no decimal
A hollow aluminum alloy [G = 3,800 ksi] shaft having a length of 12 ft, an outside diameter of 4.50 in., and a wall thickness of 0.50 in. rotates at 3 Hz. The allowable shear stress is 6 ksi, and the allowable angle of twist is 5°. What horsepower may the shaft transmit?
Answer:
Horse power = 167.84 hp
Explanation:
Horsepower is calculated using the formula;
P = T * w
See the attached file for the calculation
Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find the Fourier series coefficient of x(t). Then find its Fourier transform and sketch the magnitude and phase spectra.b)Sketch the magnitude and phase spectra of the output (i.e., |Y(????)|and∡Y(????)) if the system is a low-pass filter with H(????)={1|????|<3????020other????ise, where ????0=2πT0.c)Sketch the magnitude and phase spectra of the output(|Y(????)|and∡Y(????)) if the system is a high-pass filter with H(????)={1|????|>5????020other????ise, where ????0=2πT0.d)Sketch the magnitude and phase spectra of the outputif the system is a filter with H(????)=11+j????.
Answer:
See explaination
Explanation:
The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by
|Y(w)| = |X(w)|
The phase spectrum of y(t) is given by
<Y(w) = -wto + <X(w)
please kindly see attachment for the step by step solution of the given problem.
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using the ductile Coulomb-Mohr theory, determine the factor of safety for the states of plane stress.
Answer:
2.135
Explanation:
Lets make use of these variables
Ox 16.5 kpsi, and Oy --14,5 kpsi
To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.
Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.
Please refer to attachment for the step by step solution.
The factor of safety for the states of plane stress can be determined using the ductile Coulomb-Mohr theory. The factor of safety is the ratio of the minimum yield strength to the maximum stress in the material. In this case, the factor of safety is approximately 0.889.
Explanation:The factor of safety for the states of plane stress can be determined using the ductile Coulomb-Mohr theory. The factor of safety is defined as the ratio of the minimum yield strength in tension and compression to the maximum stress in the material. In this case, the minimum yield strength in tension is Syt = 60 kpsi and in compression is Syc = 75 kpsi.
To find the factor of safety, we need to determine the maximum stress in the material. The maximum stress can be calculated using the formula:
Maximum stress = (Syt + Syc) / 2
Plugging in the values, we get:
Maximum stress = (60 + 75) / 2 = 67.5 kpsi
Now, we can calculate the factor of safety using the formula:
Factor of safety = Syt / Maximum stress
Plugging in the values, we get:
Factor of safety = 60 / 67.5 = 0.889
Therefore, the factor of safety for the states of plane stress is approximately 0.889.
A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=0.33) followed by 200 ft. on an unpaved shoulder (F=0.28), what is the initial speed of the vehicle just before the pavement skid was begun?
Answer:
[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]
Explanation:
The deceleration of the car on the dry pavement is found by the Newton's Law:
[tex]\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}[/tex]
Where:
[tex]a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g[/tex]
[tex]a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a_{1} = -3.040\,\frac{m}{s^{2}}[/tex]
Likewise, the deceleration of the car on the unpaved shoulder is:
[tex]a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g[/tex]
[tex]a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a_{2} = -2.549\,\frac{m}{s^{2}}[/tex]
The speed just before the car entered the unpaved shoulder is:
[tex]v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}[/tex]
[tex]v_{o} = 18.175\,\frac{m}{s}[/tex]
And, the speed just before the pavement skid was begun is:
[tex]v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}[/tex]
[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]
The initial speed of the vehicle just before the pavement skid was begun is 5284.65 ft/hr.
Dry pavement friction coefficient Fdry = 0.33
Length of skid marks on dry pavement ddry = 100 ft
Friction coefficient on unpaved shoulder Fshoulder = 0.28
Length of skid marks on unpaved shoulder = 200 ft
First, let's calculate the work done on dry pavement:
Work on dry pavement = Fdry × ddry = [tex]0.33 *100[/tex]
= 33 ft·lbf
Work on unpaved shoulder = Fshoulder × dshoulder
= [tex]0.28 * 200[/tex]
= 56 ft·lbf
Total work done = Work on dry pavement + Work on unpaved shoulder = 33 + 56
= 89 ft·lbf
Assuming the car's mass remains constant, and the final speed is 0, we have:
89 ft·lbf = (1/2)m × (10 mi/hr)²
Convert the final speed to feet per hour:
[tex]10 mi/hr = 10 × 5280/3600 = 5280 ft/hr[/tex]
Now, solve for the initial speed:
v = √((2 × 89 ft·lbf) / m)
v ≈ √((2 × 89) / (6.38 × 10⁻⁶)) ft/hr
v ≈ [tex]\sqrt{(27889055.9}[/tex] ft/hr
v ≈ 5284.65 ft/hr
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the internal pressure is 60 psi, determine the maximum normal stress and the maximum shearing stress in the container.
Answer:
Maximum Normal Stress σ = 8.16 Ksi
Maximum Shearing Stress τ = 4.08 Ksi
Explanation:
Outer diameter of spherical container D = 17 ft
Convert feet to inches D = 17 x 12 in = 204 inches
Wall thickness t = 0.375 in
Internal Pressure P = 60 Psi
Maximum Normal Stress σ = PD / 4t
σ = PD / 4t
σ = (60 psi x 204 in) / (4 x 0.375 in)
σ = 12,240 / 1.5
σ = 8,160 P/in
σ = 8.16 Ksi
Maximum Shearing Stress τ = PD / 8t
τ = PD / 8t
τ = (60 psi x 204 in) / (8 x 0.375 in)
τ = 12,240 / 3
τ = 4,080 P/in
τ = 4.08 Ksi
A supply fan is operating at 30000 cfm and 4 inch of water with an efficiency of 50%. (a) Calculate the fan power at the current operating condition. (b) Calculate the pressure and power if the supply fan reduces its speed to deliver 20000 cfm.
Answer:
The pressure and power of fan is 1.77 and 11.18 Hp respectively.
Explanation:
Given:
Discharge [tex]Q_{1} = 30000[/tex] cfm
Pressure difference [tex]\Delta P = 4[/tex] inch
Efficiency [tex]\eta = 50\%[/tex]
(A)
From the formula of fan power,
[tex]P _{1} = \frac{Q \Delta P}{6356 \eta}[/tex]
[tex]P_{1} = \frac{30000 \times 4}{6356 \times 0.5}[/tex]
[tex]P_{1} = 37.76[/tex] Hp
(B)
Fan power and pressure is given by,
We know that pressure difference is proportional to the square of discharge.
[tex]\frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}[/tex]
[tex]\Delta P_{2} = (\frac{20000}{30000} ) \times 4[/tex]
[tex]\Delta P_{2} = 1.77[/tex]
Fan power proportional to the cube of discharge.
[tex]\frac{P_{2} }{P^{1} } = (\frac{Q_{2} }{Q_{1} } )^{3}[/tex]
[tex]P_{2} = \ (\frac{20000}{30000} ) ^{3} \times 37.76[/tex]
[tex]P_{2} = 11.18[/tex] Hp
Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.
The aerodynamic behavior of a flying insect is to be investigated in a wind tunnel using a ten-times scale model. It is known that the insect’s velocity depends on its size (characteristic length L), wing flapping frequency ω, surrounding fluid’s density rho and viscosity μ. If the insect flaps its wings 50 times a second when flying at 1.25 m/s, determine the wind tunnel air speed and wing oscillation frequency required for dynamic similarit
Answer:
Explanation:
Write the equation for Reynolds number as follows:
Re = VL/v
For dynamic similarity,
(VL/v)m + (VL/v)p…… (1)
Since, the model and prototype are in same medium, the kinematic viscosity remains same.
From equation (1), we can write
(VL)m = (VL)p
Here, L represents length, and V is the velocity.
Re-write the equation as follows:
Vm = Lp/Lm x Vp
Substitute 1/8 for Lp/Lm and 1.5m/s for Vp .
Vm = 1/8 x 1.5
Vm = 0.1875m/s
Therefore, the wind tunnel air speed is 0.1875m/s.
A simply supported beam spans 20 ft and carries a uniformly distributed dead load of 0.8 kip/ft, including the beam self-weight, and a live load of 2.3 kip/ft. Determine the minimum required plastic section modulus (Zx) and select the lightest-weight W-shape to carry the moment. Consider only the limit state of yielding and use A992 steel.
Answer:
Bending stress = 32.29ksi ∠ 33.0 ksi
Explanation:
CHECK BELOW FOR THE ANSWER IN THE FILE ATTACHED.
Steam enters the turbine of a power plant operating on the Rankine cycle at 3300 kPa and exhausts at 50 kPa. To show the effect of superheating on the performance of the cycle, calculate the thermal efficiency of the cycle and the quality of the exhaust steam from the turbine for turbine-inlet steam temperature of 450°C
Answer:
Thermal efficiency of cycle = 0.314
Quality of exhaust steam = 0.959
Explanation:
An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tensile force of 1 kN is applied axially to this heated rod, what is the expected mechanical response
Answer: the metal will experience a strain of approximately 2.037Mpa
This strain is lesser than if the force was applied at room temperature.
This will reduce internal stress and increase some mechanical properties of the aluminum such as mechanical hardening.
Explanation:
Detailed explanation and calculation and comparison with equivalent tensile stretching at ordinary room temperature is compared.
Consider an oil-to-oil double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C.
a. Do you think this is a parallel-flow or counter-flow heat exchanger? Why?
b. Assuming the mass flow rates of both fluids to be identical, determine the effectiveness of this heat exchanger.
a. This is a counter flow heat exchanger.
b. The effectiveness of heat exchanger is 0.615
A counter-flow heat exchanger is one in which the direction of the flow of one of the working fluids is opposite to the direction to the flow of the other fluid. In a parallel flow exchanger, both fluids in the heat exchanger flow in the same direction.Counter flow heat exchanger distributes the heat more evenly across the heat exchanger and allows for maximum efficiency.The effectiveness of heat exchanger is defined as ratio of actual heat transfer to the maximum possible heat transfer.
Given that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C.
[tex]Effectiveness=\frac{55-15}{80-15}=\frac{40}{65}=0.615[/tex]
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Use phasor techniques to determine the current supplied by the source given that V = 10 <0o v, R = 7 Ω, C = 18 μF, L = 4 mH and ω = 2000 rad/sec. The impedance seen by the source is Z = ∠ o Ω. (Round the magnitude to three decimal places and the angle to two decimal places.) The current supplied by the source is I = ∠ ° A. (Round the magnitude to three decimal places and the angle to two decimal places.)
Answer
The Impedance Z = 20.982 ohms
The phase angle is Φ= - 70.51°
The current I= 0.477amp
Explanation:
This problem bothers on alternating current, this time an R-L-C circuit
What is a R-L-C circuit?
An RLC circuit is an electrical circuit consisting of a resistor, an inductor, and a capacitor, connected in series or in parallel. The name of the circuit is derived from the letters that are used to denote the constituent components of this circuit.
N/B : Kindly find attached solutions and diagrams for your reference
A 1 phase load operates at 600 V and consumes 75 kW with a 0.85 lagging power factor. Compute the complex power consumed by the load. Type the answer with two decimal digits. The answer is in VA units.
Answer:
Complex power=84 W
Explanation:
using equation
s=vs^2/2z
s=600^2/2z
s=84+29.4j
using s=P+jq
complex power=P=84 W
Answer:
Complex power (s) = 88235.3costheta + 88235.3sintheta
Explanation:
Wapplied = U* I *PF / 1000
where
Wapplied = real power (kilowatts, kW)
U = voltage (volts, V) = 600V
I = current (amps, A) = ?
PF = power factor = 0.85
I = Wapplied * 1000/ U * PF
I = (75 * 1000)/(600*0.85)
I = 147.058
Power = IV
Complex power (s) = IVcos thetha + IV sinthetha
Complex power (s) = 88235.3costheta + 88235.3sintheta
A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere. For a measured manometer deflection of 5 m of water, determine the tank static pressure if the barometric pressure is 101.3 kPa absolute.
(a) 140 kPa absolute
(b) 150 kPa absolute
(c) 160 kPa absolute
(d) 170 kPa absolute
Answer:
[tex]P = 150.335\,kPa[/tex] (Option B)
Explanation:
The absolute pressure of the air-filled tank is:
[tex]P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)[/tex]
[tex]P = 150.335\,kPa[/tex]
You wish to filter out 60 Hz noise (which arises from electrical interference at the frequency of AC current in our electrical grid) using a simple RC circuit, which is one useful form has a transfer function:
G(s) = 1/RCS + 1
a. What value of the product RC should you choose so that amplitude at 60 Hz is attenuated by 90%?
b. For that value of RC, what is the largest frequency that is attenuated by less than 5%?
c. Turn in a Bode plot for your proposed system generated in Matlab.
Answer:
a. 0.02639
b. 1.98H
Explanation:
Please see attachment
An element representing maximum in-plane shear stress with the associated ________ normal stresses is oriented at an angle of _______ from an element representing the ________ stresses.
Answer: 1) tensile
2) 45 degree orientation
3) principal shear stress
Explanation:
An element representing maximum in-plane shear stress with the associated _tensile normal stresses is oriented at an angle of ____45_degree__ from an element representing the ____principal shear____ stresses.
A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surface is 5 m. The atmospheric pressure is 95 kPa. Considering the atmospheric pressure, the hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.
Answer:
864 KN
Explanation:
Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.
Please kindly check attachment for the step by step solution of the given problem.
You are designing a vascular prosthesis made of woven Dacron and you would like to reinforce it with thin metal wires. The prosthesis has a cylindrical shape. The principal directions of stress are axial and circumferential. What orientation would you choose for the metal wires to avoid rupture/delamination?
Answer:
In place of spiral configuration, we can also use metal wire rings which will provide better strength against circumferential stress. But this spiral configuration will provide strength against circumferential along with axial stress also. If the vascular prosthesis gets elongated along axial direction the corresponding radius of the rings will become short and it will prohibit the expansion of the vascular prosthesis along the circumferential direction.
Explanation:
The orientation is attached below.
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is Δp1. If the length of the pipe is then doubled, what is the relation of the new pressure drop Δp2 to the original pressure drop Δp1 at the original mass flow rate?
delta P2=?
Answer:
[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]
Explanation:
The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.
[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]
In this assignment, you will demonstrate your ability to write simple shell scripts. This is a cumulative assignment that will challenge you to pull together a variety of lessons from throughout the course. Earlier in the semester, when looking at the problems of file transfer, we discussed the difference beteween binary and ASCII files and the two styles of ASCII text files. In this assignment, you will write a script to diagnose whether a group of files are Windows-style ASCII text and, if so, will convert them to Unix-style ASCII text.
Answer:
Explanation:
Usage: flip [-t|-u|-d|-m] filename[s]
Converts ASCII files between Unix, MS-DOS/Windows, or Macintosh newline formats
Options:
-u = convert file(s) to Unix newline format (newline)
-d = convert file(s) to MS-DOS/Windows newline format (linefeed + newline)
-m = convert file(s) to Macintosh newline format (linefeed)
-t = display current file type, no file modifications
the upper surface of a 1x1 ASTM B152 copper plate is being cooled by air at 20 c. if the rate of convection heat transfer from the plate surface is 700 W, would the use of ASTM B152 plate be in compliance with tht ASME Code for process piping?
Answer:
Since the Reynolds number is below 3 x 10⁵ the use of ASTM B152 plate WILL NOT be in complaince with ASME Code for process piping.
Explanation:
Detailed Explanation is given in the attached document.
Create a shell script (utilities1.sh) that will print a menu of commands to execute. (a) The script will prompt the user for a number as input. The input will be read, the corresponding command will be executed, and you will see the output of the executed command on your screen. The script will then exit normally, and return a value of 0 (exit 0). See the sample run below.
Answer:
Explanation:
check the attached files for solution.
What evidence indicates that a reaction has occurred? (Select all that apply.)
The temperature decreased.
A solid brown product formed.
The temperature increased.
A gas formed.
An explosion occurred.
The evidence that indicates that a reaction has occurred include the following:
B. A solid brown product formed.
C. The temperature increased.
D. A gas formed.
E. An explosion occurred.
A chemical reaction is a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.
This ultimately implies that, a chemical change would give rise to the chemical properties of matter by causing the transformation of one chemical substance into one or more different chemical substances.
Derive the following formula for the complex Fourier Series representation of a full wave rectified sine wave of unit amplitude. Remember that the period of the rectified signal is half of that of the sine wave.
Answer:
Please see attachment for good equations
A 20 mm diameter cylindrical rod fabricated from a 2014-T6 alloy is subjected to repeated tension-compression load cycling along its axis. The magnitudes of the maximum tensile and compressive stresses during cycling are 600 MPa and 100 MPa, respectively. Estimate the fatigue lifetime of this cylindrical rod (in years), if it is cycled between these stress states sinusoidally with a frequency of 0.01 Hz. You may assume that the curves represent a 100% chance of failure.
Answer:
0.0317
Explanation:
Please see attachnment
Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly from the source to the depositing surface without being scattered along the way.
a. To get an idea of how few and far between the air molecules are in a thin-film deposition chamber, determine the mean free path of a generic "air" molecule with an effective diameter of 0.25 nm at a pressure of 1.5 x 10-6 Pa and temperature of 300 K.
b. If the chamber is spherical with a diameter of 10 cm, estimate how many times a given molecule will collide with the chamber before colliding with another air molecule.
c. How many air molecules are in the chamber (treating "air" as an ideal gas)?
Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
[tex]\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}[/tex]
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
[tex]\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m[/tex]
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
[tex]n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times[/tex]
c) By using the equation of the ideal gases you obtain:
[tex]PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules[/tex]
A doubly drained specimen, 2.54 cm in height, is consolidated in the lab under an applied stress. The time for 50 % overall (or average) consolidation is 12 min. (a) Compute the cv value for the lab specimen. (b) How long will it take for the specimen to consolidate to an average consolidation of 90 %? (c) If the final consolidation settlement of the specimen is expected to be 0.43 cm, how long will it take for 0.18 cm of settlement to occur? (d) After 14 minutes, what percent consolidation has occurred at the middle of the specimen?
Answer:
Cv = 0.026 cm²/min
t = 52.60 min
v% = 41.86 %
tv = 0.1375
t = 8.53 min
v = 53.61 %
Explanation:
given data
height = 2.54 cm
50 % consolidation = 12 min
solution
we get here first Cv value that is express as
Tv = [tex]\frac{Cv\times t}{d^2}[/tex] .................1
here Tv for 50% is 0.196
put here value and we get
0.196 = [tex]\frac{Cv\times 12}{\frac{2.54}{2}^2}[/tex]
solve it we get
Cv = 0.026 cm²/min
and
for tv for 90 % consolidation is 0.848
put value in equation 1
0.848 = [tex]\frac{0.026\times t}{\frac{2.54}{2}^2}[/tex]
solve it we get t
t = 52.60 min
and
v% will be here is
v% = [tex]\frac{0.18}{0.43} \times 100[/tex]
v% = 41.86 %
and
tv = [tex]\frac{\pi }{4}\times \frac{4}{100}^2[/tex]
tv = 0.1375
so now put value in equation 1 we get
0.1375 = [tex]\frac{0.026 \times t}{\frac{2.54}{2}^2}[/tex]
solve it we get
t = 8.53 min
and
now put value of t 14 min in equation 1 will be
tv = [tex]\frac{0.026 \times 14}{\frac{2.54}{2}^2}[/tex]
t = 0.225 min
and v will be after 14 min
0.0225 = [tex]\frac{\pi }{4}\times \frac{v}{100}^2[/tex]
v = 53.61 %