Answer:
Human Capital Management (HCM) will help the start-up firm manage its recruiting and hiring activities.
Explanation:
Human Capital Management (HCM) Platform will assist the start-up firm manage its main point of access by keeping the employee records and maintaining the wages and salaries, managing the benefits, time, and attendance, and carrying out performance reviews including looking after the most important asset employees.
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
A flow inside a centrifuge can be approximated by a combination of a central cylinder and a radial line source flow, giving the following potential function:
Ø= a2/r -cosØ + aßlnr = r
Where a is the radius of the central base of the centrifuge and ß is a constant.
a) Provide expressions for the velocities Vr and vo .
b) Find the expression for the stream function.
Answer:
a) Vr = - a^2/r cosθ + aß / r
Vθ = 1/r [ -a^2/r * sinθ ]
b) attached below
Explanation:
potential function
Ø= a^2 /r cosØ + aßlnr ----- ( 1 )
a = radius , ß = constant
a) Expressions for Vr and Vθ
Vr = dØ / dr ----- ( 2 )
hence expression : Vr = - a^2/r cosθ + aß / r
Vθ = 1/r dØ / dθ ------ ( 3 )
back to equation 1
dØ / dr = - a^2/r sinθ + 0 --- ( 4 )
Resolving equations 3 and 4
Vθ = 1/r [ -a^2/r * sinθ ]
b) expression for stream function
attached below
The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and the spring stretched 39 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 99 mm, and (b) calculate the distance d traveled by the block before it comes to rest.
Solution :
The spring is expanded by 2 times of the block when it moves down an inclined by x times.
Here, [tex]$x_1$[/tex] = 39 mm
[tex]x_2[/tex] = 225 mm
a). From the work energy principal,
Work forces = kinetic energy
[tex]$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$[/tex]
[tex]$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$9.75= 7.5(V^2_2-0.08^2)$[/tex]
[tex]$1.3= V^2_2-0.08^2$[/tex]
[tex]$V_2=1.14\ m/s$[/tex]
b). calculating the distance travelled by the block before it comes to rest.
Substitute the value of [tex]V_2[/tex] in (1),
[tex]$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$[/tex]
[tex]$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$[/tex]
[tex]$98.43x - 100(4x^2+0.156x)+0.048=0$[/tex]
[tex]$98.43x - 400x^2-15.6x+0.048=0$[/tex]
[tex]$82.83x - 400x^2+0.048=0$[/tex]
[tex]$ 400x^2- 82.83x-0.048=0$[/tex]
x = 0.20 m
Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.
Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
[tex]\to \frac{P}{2} \cos \theta =mg \sin \theta\\\\\to P=2 mg \tan \theta \\\\[/tex]
[tex]=2 \times 1 \times g \times \tan 37^{\circ}\\\\ =2 \times 10 \times \frac{3}{4}\\\\ =15 \ N[/tex]
A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).
Answer:
The right answer is "2.2099 m³".
Explanation:
Given:
Mass,
m = 5 kg
Temperature,
T = 35℃
or,
= 35 + 273
Pressure,
P = 200 kPa
Gas constant,
R = 0.2870 kj/kgK
By using the ideal gas equation,
The volume will be:
⇒ [tex]PV=mRT[/tex]
or,
⇒ [tex]V=\frac{mRT}{P}[/tex]
By substituting the values, we get
[tex]=\frac{5(0.2870)(35+273)}{200}[/tex]
[tex]=\frac{441.98}{200}[/tex]
[tex]=2.2099 \ m^3[/tex]
A container can be made from steel [β = 36 × 10-6 (C°)-1] or lead [β = 87 × 10-6 (C°)-1]. A liquid is poured into the container, filling it to the brim. The liquid is either water [β = 207 × 10-6 (C°)-1] or ethyl alcohol [β = 1120 × 10-6 (C°)-1]. When the full container is heated, some liquid spills out. To keep the overflow to a minimum, from what should the container be made and what should the liquid be
Answer:
The container should be made of lead and the liquid should be water.
Explanation:
Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.
So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container
For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)
So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]
Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum
where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.
So, trying each combination, with β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹
With β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹
The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
Since β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹] = coefficient of expansion for water, the container should be made of lead and the liquid should be water.
WILL MARK BRAINLIST I need help on this asap thanks
Determine the dimensions for T if T = M V^2 A / L^3 where M is a mass, V is a velocity, A is an area, and L is a length.
L / T
M
M L / T^2
M / (L T^2)
No dimensions
Explanation:
ask your dad please and use your brain
Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.
Answer:
you are right but then you ddnt ask a question
The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
Answer:
The explanation according to the given query is summarized in the explanation segment below.
Explanation:
If somehow the fin has become too lengthy, this same fin tip temperature approaches the temperature gradient and maybe we'll ignore heat transmission out from end tips.Additionally, effective heat transmission as well from the tip could be ignored unless the end tip surface is relatively tiny throughout comparison to its overall surface.what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
Represent each of the following units as a combination of primitive
dimensions where M=mass, L=length, T=time. As an example, miles per hour would
correspond to [L/T].
a. kilometer
b. quart
c. pascal
d. watt
e. newton
f. horsepower
Answer:
a. unit of length: [L]
b. unit of volume: [[tex]L^3[/tex]]
c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]
d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]
f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]
Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]
Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]
where:
F = force
A = area
W = work
t = time
a = acceleration
v = velocity
x = displacement
Select the correct statement(s) regarding network physical and logical topologies.
a. While logical topologies can be configured in star, ring, bus, and tree configurations, the physical topology must always be in a full-mesh topology
b. logical topologies always incorporate centralized access, whereas physical topologies are always configured as a distributed access network
c. the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another
d. all statements are correct
Answer:
The physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another ( C )
Explanation:
Network physical is simply the process/method of connecting the Network using cables while Logical topology is the general architecture of the communication mechanism in the network for all nodes.
Hence The correct statement is the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape until the tank pressure drops to 200 kPa. Assuming that the argon remaining in the tank experiences a reversible adiabatic process, find the final mass of argon in the tank. Since you don't have argon gas tables, assume cp, cv, k as needed at some appropriate temperature(s).
Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
Calculate the value of the M2
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
Note: Calculation for T2 is attached below
A work-mode-choice model is developed from data acquired in the field in order to determine the probabilities of individual travelers selecting various modes. The mode choices include automobile drive-alone (DL), automobile shared-ride (SR), and bus (B). The utility functions are estimated as follows:
UDL = 2.6 - 0.3 (costDL) - 0.02 (travel timeDL)
USR = 0.7 - 0.3 (costSR) - 0.04 (travel timeSR)
UB = -0.3 (costB) - 0.01 (travel timeB)
where cost is in dollars and time is in minutes. The cost of driving an automobile is exist5.50 with a travel time of 21 minutes, while the bus fare is exist1.25 with a travel time of 27 minutes. How many people will use the shared-ride mode from a community of 4500 workers, assuming the shared-ride option always consists of three individuals sharing costs equally?
a. 314
b. 828
c. 866
d. 2805
Answer:
b. 828
Explanation:
UDL = 2.6 - 0.3 [5.5] - 0.02 [ 21 ] = 0.53
USR = 0.7 - 0.3 [5.5 / 2 ] - 0.04 [ 21 ] = -0.69
UB = -0.3 [ 1.25 ] - 0.01 = -0.645
Psr = [tex]\frac{e^{-0.53} }{e^{-0.53} + e^{-0.69} + e^{-0.645} }[/tex]
Solving the equation we get 0.184.
Number of people who will take shared ride is:
0.184 * 4500 = 828 approximately.
how many types of lavatory there is?
Answer:
there are generally two types of toilet bowl types- round and elongated.
Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Output her commission for that month. Note that you will need the following Variables: SalesAmount CommissionRate CommissionEarned
You will need the following formula: CommissionEarned= Sales Amount * (commissionrate/100)
Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
Match the test to the property it measures.
a. Rockwell
b. Inston
c. Charpy
d. Fatigue
e. Brinell
f. Izod
1. impact strength
2. stress vs strain
3. hardness
4. Endurance Limit
Answer:
a. Rockwell 3. hardness
b. Instron 2. stress vs strain
c. Charpy 1. impact strength
d. Fatigue 4. Endurance Limit
e. Brinell 3. hardness
f. Izod 1. impact strength
Explanation:
Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.
Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.
Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.
Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.
What must you do to become ASE certified as an automotive technician?
Answer:
To become ASE certified, you must pass an ASE test and have relevant hands-on work experience. The amount of work experience required can vary by test, and is specified in detail here. ASE recommends submitting the form after you've registered to take an ASE certification test.
Good luck!
Explanation:
Answer: One theme in White Fang is adapting in order to survive. White Fang finally submits to Gray Beaver. He also copes with fighting other dogs. White Fang changes his behaviors so that he can live.
Explanation: its the sample response
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
A horizontal water jet impinges against a vertical flat plate at 30 ft/s and splashes off the sides in the verti- cal plane. If a horizontal force of 500 lbf is required to hold the plate against the water stream, determine the volume flow rate of the water.
Answer:
8.6 ft³/s
Explanation:
The force due to the water jet F = mv where m = mass flow rate = ρQ where ρ = density of water = 62.4 lbm/ft³ and Q = volume flow rate. v = velocity of water jet = 30 ft/s
So, F = mv
F = ρQv
making Q subject of the formula, we have
Q = F/ρv
Since F = force due to water jet = force needed to hold the plate against the water stream = 500 lbf = 500 × 1 lbf = 500 × 32.2 lbmft/s² = 16100 lbmft/s²
Since
Q = F/ρv
Substituting the values of the variables into the equation for Q, we have
Q = F/ρv
Q = 16100 lbmft/s²/(62.4 lbm/ft³ × 30 ft/s)
Q = 16100 lbmft/s²/1872 lbm/ft²s
Q = 8.6 ft³/s
So, the volume flow rate is 8.6 ft³/s.
Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbine at a pressure of 125 kPa. At steady state, the work output of the turbine is 530 kJ/kg. The surrounding air is at 20 o C. Neglect the changes in kinetic energy and potential energy. Determine (20 points) (a) the heat transfer from the turbine to the surroundings per unit mass flow rate, (b) the entropy generation during this process.
Answer:
a) -505.229 kJ/Kg
b) -1.724 kJ/kg
Explanation:
T1 = 400°C
P1 = 3 MPa
P2 = 125 kPa
work output = 530 kJ/kg
surrounding temperature = 20°C = 293 k
A) Calculate heat transfer from Turbine to surroundings
Q = h2 + w - h1
h ( enthalpy )
h1 = 3231.229 kj/kg
enthalpy at P2
h2 = hg = 2676 kj/kg
back to equation 1
Q = 2676 + 50 - 3231.229 = -505.229 kJ/Kg ( i.e. heat is lost )
b) Entropy generation
entropy generation = Δs ( surrounding ) + Δs(system)
= - 505.229 / 293 + 0
= -1.724 kJ/kg
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
3-71A 20mm diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15 degrees, what must be the length of the bar
Answer:
The right answer is "1.903 m".
Explanation:
Given that,
[tex]\tau =110 \ MPa[/tex]
[tex]G=80 \ GPa[/tex]
[tex]\Theta=15\times \frac{\pi}{180}[/tex]
[tex]=\frac{\pi}{12}[/tex]
[tex]d=20 \ mm[/tex]
As we know,
⇒ [tex]\frac{\tau}{r}=\frac{G \Theta}{L}[/tex]
Or,
⇒ [tex]L=\frac{G \theta r}{\tau}[/tex]
[tex]=\frac{80\times 10^3}{110}\times \frac{\pi}{12}\times 10[/tex]
[tex]=1903.9 \ mm[/tex]
or,
[tex]=1.903 \ m[/tex]
6. When the engine stalls or the power unit fails, on a car with power
brakes, the service brake pedal will
A. Take about the same amount of pressure
B. Take more pressure to stop
C. Take less pressure to stop
D. Become locked in place and no longer help stop the car
HELP PLEASE!! ASAP!!!!
can some answer this 2 questions please as paragraph i want it nowww it is graded what action should be taken to make it safe ? also the first question
Actions violated:
Long hair isn't tied upThe girl isn't wearing a lab coatThe girl isn't wearing safety gogglesExtra: There doesn't seem to be an emergency fire blanket in the safeActions to be taken:
Make sure the girl wears a lab coat or kick her outMake sure the girl wears safety goggles or kick her outMake sure her hair is tied up or kick her outEdit: Use these to write your paragraph.
The following measurements are taken on particular junction diodes for which V is the terminal voltage and I is the diode current. For each diode, estimate values of Is and the terminal voltage at 10% of the measured current.
(a) V = 0.700 V at I = 1.00 A.
(b) V = 0.650 V at I = 1.00 mA.
(c) V = 0.650 V at I = 10 mu A.
(d) V = 0.700V at I = 100 mA.
The values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.
The relation between the current and voltage of a diode is given by the Shockley diode equation. It is an exponential function and can be given by the following equation:
[tex]I = Is \times (e^{V/Vt} - 1)[/tex]
Where
I = currentV = voltageVt = thermal voltageIs = reverse saturation current.(a)
Given that:
V = 0.700 V
And I = 1.00 A.
Substituting these values in the equation above to get,
[tex]1.00 A = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 A in the above equation and solving for V as:
V = 0.581 V.
(b)
Given that:
V = 0.650 V
And, I = 1.00 mA.
Substituting these values in the equation above to get,
[tex]1.00 mA = Is \times (e^{0.650 V / 0.025 V} - 1)\\ Is = 4.56 \times 10^{-15} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 mA in the above equation and solving for V as:
V = 0.516 V.
(c)
Given that:
V = 0.650 V
And, I = 10 μA.
Substituting these values in the equation above to get,
[tex]10 \mu A = Is \times (e^{0.650 V / 0.025 V} - 1)\\Is = 1.18 \times 10^{-16} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 1 μA in the above equation and solving for V as:
V = 0.459 V.
(d)
Given that:
V = 0.700 V
And, I = 100 mA.
Substituting these values in the equation above to get,
[tex]100 \ mA = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]
The terminal voltage at 10% of the measured current can be found by substituting I = 10 mA in the above equation and solving for V as:
V = 0.581 V.
So, the values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.
Learn more about Terminal voltage here:
https://brainly.com/question/34372613
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In an international film festival, a penal of 11 judges is formed to judge the best film. At
last two films FA and FB were considered to be the best where the opinion of judges got
divided. Six judges where in favor of FA whereas five in favor of FB. A random sample
of five judges was drawn from the panel. Find the probability that out of five judges,
three are in favor of film FA.Enunciate demerits of classical probability.
Answer:
International Film Festival
Judging the best best film:
a. The probability that out of five judges (random sample), three are in favor of film FA is:
= 33%.
b. The demerits of classical probability are:
1. Classical probability can only be used with events that have definite numbers of possible outcomes.
2. Classical probability can only handle events where each outcome is equally likely.
3. Classical probability is based on the assumption of linear relationship (which is not always true in real life) between the latent variable and observed scores.
Explanation:
a) Number of judges = 11
Number of judges in favor of FA film = 6
Number of judges in favor of FB film = 5
Probability of judges in favor of FA film = 6/11
Probability of judges in favor of FB film = 5/11
Random sample of judges = 5
Probability that out of five judges, three are in favor of film FA = 3/5 * 6/11
= 18/55
= 33%
b) Classical probability is the simple probability showing that each event has equal chance of happening. It can be contrasted with empirical probability that is obtained from experiments.
In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false
Answer:
Hence the given statement is false.
Explanation:
For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.
The speed within the test section is decided by the planning of the tunnel.
Thus by adjusting the pressure difference won't change the worth of test section velocity.
Answer:
The given statement is false .
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW of energy to the water and the head loss of the flow is 10 m. Determine the velocity of the water leaving the pump and discharging into tank B.
Complete Question
Complete Question is attached below.
Answer:
[tex]V'=5m/s[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=0.10m[/tex]
Power [tex]P=4.0kW[/tex]
Head loss [tex]\mu=10m[/tex]
[tex]\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu[/tex]
[tex]\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10[/tex]
[tex]H_m=(\frac{200*10^3}{1000*9.8}-10)[/tex]
[tex]H_m=10.39m[/tex]
Generally the equation for Power is mathematically given by
[tex]P=\rho gQH_m[/tex]
Therefore
[tex]Q=\frac{P}{\rho g H_m}[/tex]
[tex]Q=\frac{4*10^4}{1000*9.81*10.9}[/tex]
[tex]Q=0.03935m^3/sec[/tex]
Since
[tex]Q=AV'[/tex]
Where
[tex]A=\pi r^2\\A=3.142 (0.05)^2[/tex]
[tex]A=7.85*10^{-3}[/tex]
Therefore
[tex]V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}[/tex]
[tex]V'=5m/s[/tex]
it is a small sharp and printed item for fine worker in trimming scallops clipping threads and cutting large eyelets
Answer:
embroidery scissor
Explanation:
is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.
hope this helps