Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.
Bank loan facilitator is a consumer service facilitator who ask and provide people loan in emergency, for the purpose of education, treatment, family events, and for other reasons. For bank loan facilitator the professional organizations should be banking and finance sector. The purpose of these organizations is to help people in financial matter seeking benefit by getting interest from customers. The requirements for joining of the employee must include strong convincing power for the employee, time management, strong and tactful communication skills. Benefits of membership of the customers can help them to seek loans on need basis on lower interest. Hospital emergency care specialist provides help to the staff and the customers in medical emergency. These professionals are necessary for the hospital, clinics, and rehabilitation centers. Purpose of the organization is to provide medical care to the patients. The requirements for joining of the employee includes ability to give information to patients and staff during emergency conditions, facilitating ambulance to rescue patients from their homes, and from other areas, providing medicine, medical equipment, and other facilities to the patients and other medical staff necessary for treatment. Benefits of membership in clinical or hospital settings can help the patient in frequent visits for treatment, concession in laboratory tests, and medication.Learn more about customer:
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(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator together with a carrier at 100 kHz, with only the upper sideband retained. In the coherent detector used to recover the local oscillator supplies a sinusoidal wave of frequency 100.02 kHz. Determine the frequency components of the detector output. (b) Repeat your analysis, assuming that only the lower sideband is transmitted.
Answer:
Explanation:
The frequency components in the message signal are
f1 = 100Hz, f2 = 200Hz and f3 = 400Hz
When amplitude modulated with a carrier signal of frequency fc = 100kHz
Generates the following frequency components
Lower side band
[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]
Carrier frequency 100kHz
Upper side band
[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]
At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be
[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]
A) The frequency Components of the Detector Output are;
80 Hz, 120 Hz and 380 Hz
B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz
Message SignalsA) We are given the frequency components in the message signal as;
f1 = 100Hzf2 = 200Hzf3 = 400HzWe are told that the carrier signal has a frequency; fc = 100kHz
Thus, the frequency components generated are;
Lower side band:
100 kHz - 100 Hz = 99.9 kHz100 kHz - 200 Hz = 99.8 kHz100 kHz - 400 Hz = 99.6 kHzUpper side band:
100 kHz + 100 Hz = 100.1 kHz100 kHz + 200 Hz = 100.2 kHz100 kHz + 400 Hz = 100.4 kHzWe are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.
Thus, the output frequencies are;
100.02 kHz - 100.1 kHz = 80 Hz
100.02 kHz - 100.2 kHz = 180 Hz
100.02 kHz - 100.4 kHz = 380 Hz
B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;
100.02 kHz - 99.9 kHz = 120 Hz
100.02 kHz - 99.8 kHz = 220 Hz
100.02 kHz - 99.6 kHz = 420 Hz
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What's the "most common" concern with using variable frequency drives (VFDs)? 1) carrier frequency 2) harmonic distortion 3) hertz modulation
The common" concern with using variable frequency drives (VFDs) is C. hertz modulation.
What is variable frequency drive?It should be noted that a variable frequency drive simply means a type of motor drive that us used in mechanical drive system.
In this case, common" concern with using variable frequency drives (VFDs) is hertz modulation
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You are tasked with designing a thin-walled vessel to contain a pressurized gas. You are given the parameters that the inner diameter of the tank will be 60 inches and the tank wall thickness will be 5/8" (0.625 inches). The allowable circumferential (hoop) stress and longitudinal stresses cannot exceed 30 ksi.
(1) What is the maximum pressure that can be applied within the tank before failure? = psi(2) If you had the opportunity to construct a spherical tank having an inside diameter of 60 inches and a wall thickness of 5/8" (instead of the thin-walled cylindrical tank as described above), what is the maximum pressure that can be applied to the spherical tank? = psi
Answer:
Explanation:
For cylinder
Diameter d = 60 inches
thickness t = 0.625 inches
circumferential (hoop) stress = 30 ksi
[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]
[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]
Therefore maximum pressure without failure is P₁ = 625 psi
ii) For Sphere
[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]
A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected in series in order that the same current shall be supplied from 240 V, 50 Hz mains. Ignore the resistance of the inductor and calculate: i. the inductance of the inductor; ii. the impedance of the circuit; iii. the phase difference between the current and the applied voltage.
Answer:
(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°
Explanation:
Solution
Given that:
I= 5 A
V = 125V
Resistance R= Not known yet
Thus
To find the resistance we have the following formula which is shown below:
R = V/I
=125/15
R =8.333Ω
Now,
Voltage = 240
Frequency = 50Hz
Current (I) remain at = 15A
Z= not known (impedance)
so,
To find the impedance we have the formula which is shown below:
Z = V/I =240/15
Z= 16Ω⇒ Z = R + jXL
Z = 8.333 + jXL = 16
Thus
√8.333² + XL² = 16²
8.333² + XL² = 16²
XL² = 186.561
XL = 13.658Ω
Now
We find the inductance of the Inductor and the impedance of the circuit.
(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:
L = XL/w
=13.658/ 2π * 50
=13.658/314.15 = 0.043 = 43.43 mH
Note: w= 2πf
(ii) For the impedance of the circuit we have the following:
z = 8.333 + j 13.658
z = 16∠58.61° Ω
(iii) The next step is to find the phase difference between the applied voltage and current.
Q = this is the voltage across the inductor in a series of resonant circuit.
Q can also be called the applied voltage
Thus,
Q is described as an Impedance angle
Therefore, Q = 58.81°
Sensors are used to monitor the pressure and the temperature of a chemical solution stored in a vat. The circuitry for each sensor produces a HIGH voltage when a specified maximum value is exceeded. An alarm requiring a LOW voltage input must be activated when either the pressure or the temperature is excessive. Design a circuit for this application
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical shape. Initially, the balloon is filled up with air at the pressure and temperature of 100 kPa and 27°C respectively and the initial diameter (D) of the balloon is 10 m. Then the balloon is heated up to the point that the volume is 1.2 times greater than the original volume (V2 =1.2V1 ). Due to elastic material used in this balloon, the inside pressure ( P ) is proportional to balloonâs diameter, i.e. P = ð¼D, where ð¼ is a constant.
Required:
a. Show that the process is polytropic (i.e. PV" = Constant) and find the exponent n and the constant.
b. Find the temperature at the end of the process by assuming air to be ideal gas.
c. Find the total amount of work that is done by the balloon's boundaries and the fraction of this work that is done on the surrounding atmospheric air at the pressure of 100 kPa.
Answer:
a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]
[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]
We have;
[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]
[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]
[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]
p₂ = 100000/0.941 = 106.265 kPa
[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
Consider a series RC circuit at the left where C = 6 µ F, R = 2 MΩ, and ε = 20 V. You close the switch at t = 0. Find (a) the time constant for the circuit, (b) the half-life of the circuit, (c) the current at t = 0, (d) the voltage across the capacitor at t = 0, and (e) the voltage across the resistor after a very long time.
Answer:
(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0
Explanation:
Solution
Given that:
C = 6 µ which is = 6 * 10^ ⁻6
R = 2 MΩ, which is = 2 * 10^ 6
ε = 20 V
(a) When it is at the time constant we have the following:
λ = CR
= 6 * 10^ ⁻6 * 2 * 10^ 6
λ =12 seconds
(b) We solve for the half life of the circuit which is given below:
d₀ = d₀ [ 1- e ^ ⁺t/CR
d = decay mode]
d₀/2 = d₀ 1- e ^ ⁺t/12
2^⁻1 = e ^ ⁺t/12
Thus
t/12 ln 2
t = 12 * ln 2
t = 12 * 0.693
t = 8.31 seconds
(c) We find the current at t = 0
So,
I = d₀/dt
I = d₀/dt e ^ ⁺t/CR
= CE/CR e ^ ⁺t/CR
E/R e ^ ⁺t/CR
Thus,
at t = 0
I E/R = 20/ 2 * 10^ 6
= 10µ A
(d) We find the voltage across the capacitor at t = 0 which is shown below:
V = IR
= 10 * 10^ ⁻6 * 2 * 10^ 6
V = 20 V
(e) We solve for he voltage across the resistor.
At t = 0
I = 0
V =0
A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet.
(a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?
Answer:
Explanation:
a ) for transformer which steps down voltage , if V₁ and V₂ be voltage of primary and secondary coil and n₁ and n₂ be the no of turns of wire in them
V₁ /V₂ = n₁ / n₂
Here V₁ = 220 V , V₂ = 5V , n₁ = 2200 n₂ = ?
220 /5 = 2200 / n₂
n₂ = 2200 x 5 / 220
= 50
b )
for 100 % efficiency
input power = output power
V₁ I₁ = V₂I₂
I₁ and I₂ are current in primary and secondary coil
220 x I₁ = 5 x .5
I₁ = .01136 A .
c )
If n₂ = 100
V₁ /V₂ = n₁ / n₂
220 / V₂ = 2200 / 100
V₂ = 10 V
V₁ I₁ = V₂I₂
220 x .01136 = 10 I₂
I₂ = .25 A.
two opposite poles repel each other
Answer:
South Pole and South Pole or North Pole and North Pole.
A float valve, regulating the flow of water into a reservoir, is shown in the figure. The spherical float (half of the sphere is submerged) is 0.1553 m in diameter. AOB is the weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which flows into the reservoir. The link is mounted on a frictionless hinge at O, and the angle AOB is 135o. The length of OA is 253 mm and the distance between the center of the float and the hinge is 553 mm. When the flow is stopped AO will be vertical. The valve is to be pressed on to the seat with a force of 10,53 N to be completely stop the flow in the reservoir. It was observed that the flow of water is stopped, when the free surface of water in the reservoir is 353 mm below the hinge at O. Determine the weight of the float sphere.
Answer:
9.29 N . . . . weight of 0.948 kg sphere
Explanation:
The sum of torques on the link BOA is zero, so we have ...
(right force at A)(OA) = (up force at B)(OB·sin(135°))
Solving for the force at B, we have ...
up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N
This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.
∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg
__
The center of the sphere of diameter 0.1553 m is below the waterline by ...
(553 mm)cos(45°) -(353 mm) = 38.0300 mm
The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...
V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L
So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...
1.64336 kg -0.695205 kg ≈ 0.948 kg
The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N
The weight of the 0.948 kg float sphere is about 9.29 N.
Talc and graphite are two of the lowest minerals on the hardness scale. They are also described by terms like greasy or soapy. Both have a crystal structure characterized by sheet-structures at the atomic level, yet they don't behave like micas. What accounts for their unusual physical properties
Answer:
The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.
Explanation:
Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.
Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.
While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.
So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.
An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of these c. A transresistance amplifier d. A current amplifier e. A transconductance amplifier Clear my choice
Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
Voltage amplifier needs high input and low output resistance.Current amplifier needs Low Input and High Output resistance.Trans-conductance amplifier Low Input and High Output resistance.Trans-Resistance amplifier requires High Input and Low output resistance.Therefore, the correct answer is "None of these "
In a hydroelectric power plant, water enters the turbine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity (m/s) to which water can be accelerated by the nozzles before striking the turbine blades.
Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
Answer:
[tex]V2 = 37.417ms^{-1}[/tex]
Explanation:
Given the following data;
Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.
Nozzle outlets = 100kPa = 100000pa.
Density of water = 1000kg/m³.
We would apply, the Bernoulli equation between the inlet and outlet;
[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]
Where, V1 is approximately equal to zero(0).
Z[tex]z_{1} = z_{2}[/tex]
Therefore, to find the maximum velocity, V2;
[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]
[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]
[tex]V2 = \sqrt{2(800-100)}[/tex]
[tex]V2 = \sqrt{2(700)}[/tex]
[tex]V2 = \sqrt{1400}[/tex]
[tex]V2 = 37.417ms^{-1}[/tex]
Hence, the maximum velocity, V2 is 37.417m/s
Consider a classroom for 56 students and one instructor, each generating heat at a rate of 100 W. Lighting is provided by 18 fluorescent lightbulbs, 40 W each, and the ballasts consume an additional 10 percent. Determine the rate of internal heat generation in this classroom when it is fully occupied. The rate of internal heat generation in this classroom when it is fully occupied is W.
Answer:
What is the probability of selecting the 4 of spade or black diamond from a deck of 52 playing cards?
a) 2/52
b) 4/52
c) 3/52
d) 1/5
Explanation:
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Complete Question:
Grain diameter 1 (mm) = 4.4E-02
Yield stress 1 (MPa) = 131
Grain diameter 2 (mm) = 7.7E-03
Yield Stress 2 (MPa) = 268
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Answer:
d = 1.3 * 10⁻² m
Explanation:
According to the Hall Petch equation:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
At [tex]d_{1} = 4.4 * 10^{-2} mm[/tex], [tex]\sigma_{y1} = 131 MPa = 131 N/ mm^2[/tex]
[tex]131 = \sigma_0 + k/\sqrt{4.4 * 10^{-2}} \\k = 27.45 - 0.2096 \sigma_0[/tex]
At [tex]d_{2} = 7.7 * 10^{-3} mm[/tex], [tex]\sigma_{y2} = 131 MPa = 268 N/ mm^2[/tex]
[tex]268 = \sigma_0 + (27.45 - 0.2096 \sigma_0)/\sqrt{7.7 * 10^{-3}} \\23.5036 = 27.47 - 0.1219 \sigma_0\\ \sigma_0 = 32.45 N/mm^2[/tex]
k = 27.45 - 0.2096(32.45)
k = 20.64
At [tex]\sigma_y = 217 MPa[/tex], reapplying Hall Petch law:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
[tex]217 =32.45 + 20.64/\sqrt{d} \\217 - 32.45 = 20.64/\sqrt{d}\\184.55 = 20.64/ \sqrt{d} \\\sqrt{d} = 20.64/184.55\\\sqrt{d} = 0.11184\\d = 0.013 mm[/tex]
d = 1.3 * 10⁻² m
Increase the sampling time by a factor of 10 (to 0.1 seconds), keeping the frequency of the square wave the same, and observe the delay. Discuss relationship between sampling time and delay from one board to another.
Answer:
Time delay increases
Explanation:
Time delay is the delay between occurance of signal. If sampling time that is time between two samples is increased, the delay in the occurance of regenerated samples is also increased.
Under the normal sign convention, the distributed load on a beam is equal to the:_______A. The rate of change of the bending moment with respect to the shear force. B. The second derivative of the bending moment with respect to the length of the beam. C. The rate of change of the bending moment with respect to the length of the beam. D. Negative of the rate of change of the shear force with respect to the length of the beam.
Answer:
Under the normal sign convention, the distributed load on a beam is equal to the: O The second derivative of the bending moment with respect to the length of the beam O Negative of the rate of change of the shear force with respect to the length of the beam.
Sorry if the answer is wrong
list everything wrong with 2020
Answer:
George Floyd (BLACK LIFES MATTER)
C O V I D - 19
Quarantine
no sports
wearing a mask
and a whole lot of other stuff
Explanation:
While having a discussion about O-rings at the bottom of filters, Technician A says that the Automotive Filter Manufacturers Council recommends that the filter O-ring be lubricated with oil after installing the filter. Technician B says that the filter O-ring should be lubricated before installation. Who is correct
Answer:
Technician B is correct
Explanation:
O- rings are used with oil transmission filters to avoid transmission failures but some people use lip seals as well. either of them is inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.
0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.
Scheduling can best be defined as the process used to determine:
Answer:
Overall project duration
Explanation:
Scheduling can best be defined as the process used to determine a overall project duration.
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa. After being compressed, the air is at 450 °C. Determine
(a) the final pressure in [MPa],
(b) the increase in total internal energy in [kJ], and
(c) the total work required in [kJ].
Note that for air R-287 J/kg.K and c.-716.5 J/kg.K, and ?-
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material. For wires of diameter D = 125 m and a convection coefficient of h = 700 W/m^2 K, determine the minimum separation distance between the two legs of the sting, L=L1+L2, to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and aluminel wires. Evaluate the thermal conductivity of copper and constantan at T300 K. Use kCh =19 W/mK and kA = l29 W/mK for the thermal conductivities of the chromel and alumel wires, respectively.
Answer:
minimum separation distance between the two legs of the sting L = L 1 + L 2 therefore L = 9.48 + 4.68 = 14.16 mL = 1.14 mExplanation:
D ( diameter ) = 125 m
convection coefficient of h = 700 W/m^2
Calculate THE CROSS SECTIONAL AREA
Ac = [tex]\frac{\pi }{4} * D^2[/tex] = [tex]\frac{\pi }{4} * ( 125 )^2[/tex] = 0.79 * 15625 = 12343.75 m^2
perimeter
p = [tex]\pi * D[/tex] = 3.14 * 125 = 392.5 m
at 300k temperature the thermal conductivity of copper and constantan from the thermodynamic property table are :
Kcu = 401 w/m.k
Kconstantan = 23 W/m.k
To calculate the length of copper wire of the thermocouple junction
L 1 = 4.6 ([tex]\frac{Kcv Ac}{h P}[/tex]) ^ 1/2 = 4.6 [tex](\frac{401 *12343.75 }{700 *392.5})^\frac{1}{2}[/tex]
L 1 = 4.6 ( 4949843.75 / 274750 )^1/2
L 1 = 9.48 m
calculate length of constantan wire
L 2 = 4.6 [tex](\frac{kcons*Ac}{hp} )^\frac{1}{2}[/tex]
= 4.6 ( (23 * 12343.75) / ( 700 * 392.5) ) ^1/2
L 2 = 4.6 ( 283906.25 / 274750 ) ^ 1/2
L 2 = 4.68 m
I) therefore the minimum separation distance between the two legs of the sting L = L 1 + L 2
L = 9.48 + 4.68 = 14.16 m
ii) Evaluating the thermal conductivity of copper and constantan
Kc ( thermal conductivity of chromel) = 19 w/m.k
Ka ( thermal conductivity of alumel ) = 29 W/m.k
distance between the legs L = L 1 + L 2
THEREFORE
L = 4.6 ( (Kcn * Ac ) / ( hp ) )^1/2 + 4.6 ( (Kac * Ac)/(hp) )^1/2
L = 4.6 [tex](\frac{Ac}{hp} )^\frac{1}{2} [ (Kcn)^\frac{1}{2} + (Kal)^\frac{1}{2} ][/tex]
L = 4.6 ( 12343.75 /( 700 * 392.5) )^1/2 * [ 19^1/2 + 29^1/2 ]
L = 4.6 ( 12343.75 / 274750 ) ^1/2 * 5.39
L = 1.14 m
The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?
Find the given attachments for complete explanation
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,500 kg/hr. The feed is known to contain174 kg of aluminum and 2,326 kg of reject. After operating for 1 hour, a total of 256 kg ofmaterials is collected in the product stream. On close inspection, it is found that 140 kg ofproduct is aluminum. Estimate the % recovery of aluminum product and the % purity of thealuminum produc
Answer:
the % recovery of aluminum product is 80.5%
the % purity of the aluminum product is 54.7%
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = 80.5%
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = 54.7%
In contouring, it is necessary to measure position and not velocity for feedback.
a. True
b. False
In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
a. True
b. False
Job shop is another term for process layout.
a. True
b. False
Airplanes are normally produced using group technology or cellular layout.
a. True
b. False
In manufacturing, value-creating time is greater than takt time.
a. True
b. False
Answer:
(1). False, (2). True, (3). False, (4). False, (5). True.
Explanation:
The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.
(1). In contouring, it is necessary to measure position and not velocity for feedback.
ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.
(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
ANSWER: a=> True.
(3). Job shop is another term for process layout.
ANSWER: b => False
JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.
(4). Airplanes are normally produced using group technology or cellular layout.
ANSWER: b => False.
(5). In manufacturing, value-creating time is greater than takt time.
ANSWER: a => True.
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core temperature is 37.0°C, and it has a surface area of 2.23 m2. What is the average thickness of its blubber? The conductivity of fatty tissue without blood is 0.20 (J/s · m · °C).
Answer:
The average thickness of the blubber is 0.077 m
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Effluents from metal-finishing plants have the potential of discharging undesirable quantities of metals, such as cadmium, nickel, lead, manganese, and chromium, in forms that are detrimental to water and air quality. A local metal-finishing plant has identified a wastewater stream that contains 5.15 wt% chromium (Cr) and devised the following approach to lowering risk and recovering the valuable metal. The wastewater stream is fed to a treatment unit that removes 95% of the chromium in the feed and recycles it to the plant. The residual liquid stream leaving the treatment unit is sent to a waste lagoon. The treatment unit has a maximum capacity of 4500 kg wastewater/h. If wastewater leaves the finishing plant at a rate higher than the capacity of the treatment unit, the excess (anything above 4500 kg/h) bypasses the unit and combines with the residual liquid leaving the unit, and the combined stream goes to the waste lagoon.
(a) Without assuming a basis of calculation, draw and label a flowchart of the process. (b) Waste water leaves the finishing plant at a rate m_ 1 ? 6000 kg/h. Calculate the flow rate of liquid to
the waste lagoon, m_ 6?kg/h?, and the mass fraction of Cr in this liquid, x6(kg Cr/kg). (c) Calculate the flow rate of the liquid to the waste lagoon and the mass fraction of Cr in this liquid for m_1 varying from 1000 kg/h to 10,000 kg/h in 1000 kg/h increments. Generate a plot of x6 versus m_ 1 .
(Suggestion: Use a spreadsheet for these calculations.) (d) The company has hired you as a consultant to help them determine whether or not to add capacity to the treatment unit to increase the recovery of chromium. What would you need to know to make this determination? (e) What concerns might need to be addressed regarding the waste lagoon?
Answer:
Explanation:
The solution of all the four parts is provided in the attached figures
Describe with an example how corroded structures can lead to environment pollution?
A piston-cylinder assembly contains 5kg of water that undergoes a series of processes to form a thermodynamic cycle. Process 1à 2: Constant pressure cooling from p1=20bar and T1=360°C to saturated vapor Process 2à 3: Constant volume cooling to p3=5 bar Process 3à 4: Constant pressure heating Process 4à 1: Polytropic process following Pv =constant back to the initial state Kinetic and potential energy effects are negligible. Calculate the net work for the cycle in kJ.
Answer:
[tex]W_{net} = - 1223 kJ[/tex]
Explanation:
State 1:
[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]
State 2:
[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]
State 3:
[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]
[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]
State 4:
[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]
Calculate the network done for the cycle
[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]
: Explain why testing can only detect the presence of errors, not their absence?
Answer:
The goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.
Explanation: