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 temperature gradient in a spherical (or cylindrical) wall at steady state will always decrease (in magnitude) with increasing distance from the center (line), i.e. radial distance.
A. True
B. False
Answer:
True
Explanation:
Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.
We have this cylinder equation as
[T1-T2 / ln(r1-r2)]2πKL
The radial distance is r2-r1
The gradient of temperature is T1-T2
From the equation,
The temperature gradient has a direct and proportional relationship to radial distance
T1-T2 ∝ ln(r2-r1)
1/T1-T2 = k(r2-r1)
This inverse relationship above confirms that the statement is true
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]
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:
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A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length
Answer:
1.49 mm
Explanation:
The modulus of elasticity, Y = stress/strain = σ/ε
σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.
ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,
So, Y = σ/ε
Y = F/A ÷ ΔL/L
Y = FL/AΔL
making the change in length ΔL subject of the formula, we have
ΔL = FL/AY
substituting the value of A into the equation, we have
So, ΔL = FL/(πd²/4)Y
ΔL = 4FL/πd²Y
Since Y = 200 GPa = 200 × 10⁹ Pa
Substituting the values of the variables into the equation, we have
ΔL = 4FL/πd²Y
ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]
ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N
ΔL = 0.149 × 10⁻² m
ΔL = 1.49 × 10⁻³ m
ΔL = 1.49 mm
The change in length of the circular rod is 1.49 mm
what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
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
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.Reinforced concrete is a raw material that has always been available, but it was unappreciated by architects until the nineteenth century.
a. True
b. False
Answer: False
Explanation:
Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.
Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.
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.
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.
A 20-mm-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°, what must be the length of the bar?
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
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]
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]
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
) Please label the following statements as either True (T) or False (F). (a) In general, the greater the % of cold work, the smaller the recrystallization grain size. (b) The higher the annealing temperature, the smaller the recrystallization grain size. (c) The greater the % of cold work, the lower the recrystallization temperature.
Answer:
A. This option is true
B. This option is false
C. This option is true
Explanation:
A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.
B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.
C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.
Atmospheric pressure is 101 kPa. Pressure inside a tire is measured using a typical tire pressure gage to be 900 kPa. Find gage pressure and absolute pressure in the tire. ___________________________________________________________________
Answer:
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
Explanation:
The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:
[tex]P_{g} = 900\,kPa[/tex]
[tex]P_{abs} = P_{atm} + P_{g}[/tex]
[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]
[tex]P_{abs} = 1001\,kPa[/tex]
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Answer:
Explanation:
The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory
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%
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.
For a steel alloy it has been determined that a carburizing heat treatment of 3-h duration will raise the carbon concentration to 0.38 wt% at a point 2.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.1 mm position for an identical steel and at the same carburizing temperature.
Answer:
The right answer is "16.5 hrs".
Explanation:
Given values are:
[tex]x_1=2.6 \ mm[/tex]
[tex]t_1=3 \ hrs[/tex]
[tex]x_2=6.1 \ mm[/tex]
As we know,
⇒ [tex]\frac{x^2}{Dt}=constant[/tex]
or,
⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]
⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]
By putting the values, we get
[tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]
[tex]=5.5\times 3[/tex]
[tex]=16.5 \ hrs[/tex]
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
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
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 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
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.
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
g The inside surface of a 17 mm inner diameter tube with a 2.4 mm thick wall indicates a temperature of 46 deg C. The outside temperature is 43 deg C. The tube is 5 m long. If the tube material has a conductivity of 0.15 W/m/K, estimate the heat transfer rate through the tube wall assuming SS 1D conduction. Indicate the direction of heat transfer with a or - sign ( meaning outward and vice versa). Express your answer using two significant digits in W.
Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
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 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