Answer:
a. Composition of Vector.
Explanation:
When a bird flies in the air, it stretches its wings into air and this movement helps it move in a certain direction. This is an example of composition of vector. Air strikes the wings in opposite direction and bird wing movement helps it move against the wind.
Refrigerant-134a enters an adiabatic compressor at -30oC as a saturated vapor at a rate of 0.45 m3 /min and leaves at 900 kPa and 55oC. Determine (a) the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the rate of exergy destruction and the second-law efficiency of the compressor. Take T0
Answer:
a) 1.918 kw
b) 86.23%
c) 0.26 kw
Explanation:
Given data:
T1 = -30°C = 243 k , T0 = 27°C
using steam tables
h1 = 232.19 KJ/kg
s1 = 0.9559 Kj/Kgk
T2 = 55°C P2 = 900 kPa
Psat = 1492 kPa, h2 = 289.95 Kj/Kg, s2 = 0.9819 Kj/kgk , m = 0.0332 kg/s
a) Determine the power input to the compressor
power input = 1.918 kw
b) Determine isentropic efficiency of compressor
Isentropic efficiency = 86.23%
c) Determine rate of exergy destruction
rate = 0.26 kw
Attached below is the detailed solution of the given problems
4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape and the ground (assumed to be uniform) be if the temperature of the tape rises to 30 C. Neglect the expansion of the earth (the radius of the earth is 6.37 X 106 m)
Answer:
2102.1 m
Explanation:
Temperature at the equator = 0⁰
Radius of the earth = 6.37x10⁶
Required:
We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.
Final temperature = ∆T = 303-273 = 30
S = 11x10^-6
The clearance R = Ro*S*∆T
=6.37x10⁶x 11x10^-6x30
= 2102.1m
Or 2.102 kilometers
Thank you
I am having trouble understanding how I got these wrong on my test. Is there something I am missing with xor?
Answer:
your answer is correct
Explanation:
You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.
A heat rate of 3 kW is conducted through a section of an insulating materials of cross-sectional area 10 m^2 and thickness 2.5 cm. If the inner (hot) surface temperature is 415°C and the thermal conductivity of the material is 0.2 W/m*k , what is the outer surface temperature?
Answer: The outer surface temperature is [tex]377.5^{o}C[/tex].
Explanation:
Given: Heat = 3 kW (1 kW = 1000 W) = 3000 W
Area = 10 [tex]m^{2}[/tex]
Length = 2.5 cm (1 cm = 0.01 m) = 0.025 m
Thermal conductivity = 0.2 W/m K
Temperature (inner) = [tex]415^{o}C[/tex]
Formula used is as follows.
[tex]q = KA \frac{(t_{in} - t_{out})}{L}[/tex]
where,
K = thermal conductivity
A = area
L = length
[tex]t_{in}[/tex] = inner surface temperature
[tex]t_{out}[/tex] = outer surface temperature
Substitute the values into above formula as follows.
[tex]q = KA \frac{(t_{in} - t_{out})}{L}\\3000 W = 0.2 \times 10 \times \frac{415 - t_{out}}{0.025 m}\\t_{out} = 377.5^{o}C[/tex]
Thus, we can conclude that the outer surface temperature is [tex]377.5^{o}C[/tex].
4.3
While a train is standing still, its smoke blows 12 m/s north.
What will the resulting velocity be of the smoke relative to the train if the train
is moving at 25 m/s south?
(3)
Compute the minimum length of vertical curve that will provide 220 m stopping sight distance for a design speed of 110 km/h at the intersection of a -3.50% grade and a +2.70% grade.
i have made notes and saved it as a pdf u can take it to answer question and make ur concept good
The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m
We are given;
Stopping sight distance; S = 220 m
Design Speed; V = 110 km/h
Intersection grade 1; G1 = +2.7
Intersection Grade 2; G2 = -3.5
From the AASHTO Table attached, we can trace the value of the radius of vertical curvature for the given stopping sight distance and design speed.From the table, at S = 220 m and V = 110 km/h, we can see that;
Radius of vertical curvature; K = 74
Now, the difference in grade given is;A = G1 - G2
A = 2.7 - (-3.5)
A = 2.7 + 3.5
A = 6.2
Formula for the minimum length of vertical curve is;L = KA
Thus;
L = 74 × 6.2
L = 458.8 m
Read more about stopping sight distance at; https://brainly.com/question/2087168
Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at ft/s^2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 18 ft/s^2.
Required:
Determine the minimum distance d between the cars so as to avoid a collision.
Answer:
Explanation:
Using the kinematics equation [tex]v = v_o + a_ct[/tex] to determine the velocity of car B.
where;
[tex]v_o =[/tex] initial velocity
[tex]a_c[/tex] = constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
[tex]S = d + v_ot + \dfrac{1}{2}at^2[/tex]
Then:
[tex]v_B = 60-12t[/tex]
The distance traveled by car B in the given time (t) is expressed as:
[tex]S_B = d + 60 t - \dfrac{1}{2}(12t^2)[/tex]
For car A, the needed time (t) to come to rest is:
[tex]v_A = 60 - 18(t-0.75)[/tex]
Also, the distance traveled by car A in the given time (t) is expressed as:
[tex]S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
Relating both velocities:
[tex]v_B = v_A[/tex]
[tex]60-12t = 60 - 18(t-0.75)[/tex]
[tex]60-12t =73.5 - 18t[/tex]
[tex]60- 73.5 = - 18t+ 12t[/tex]
[tex]-13.5 =-6t[/tex]
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
[tex]S_B = S_A[/tex]
[tex]d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
[tex]d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2[/tex]
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Discuss the organizational system that you believe would be the most effective for the safety officer in a medium-sized (100-200) manufacturing facility
Answer:
A safety manager is a person who designs and maintains the safety elements at workplace. A balance should be required for production and the job in providing work environment. As a safety officer in a medium sized manufacturing facility the following organizational system can be designed and maintained:
Maintaining a workplace as per the guidelines by Occupational safety and health association. The rules and regulation should be such that maintains the manufacturing facilities. For warning to workers proper labelling, floor mapping, signs, posters should be used. Procurement and usage of safe tools. A guideline that describes safety standard and precautionary measures should be available to the workers. They should be aware about all the steps that needs to be taken in crisis. Ensuring that the workers have enough training safety and health or accident prevention. Identify and eliminate the hazardous elements from the workplace. A strict action should be taken against the worker in case of violation of rules and not adhering with guidelines.
A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I has a thermal efficiency equal to one third the thermal efficiency of R.
a. If each cycle receives the same amout of energy by heat transfer from the hot reservor, detrmine which cycle:
1. develops greater net work.
2. discharges greater energy by heat transfer to the cold reservoir
b. If each cycle develops the same net work, determine which cycle:
1. receives greater net energy by heat transfer from the hot reservoir
2. discharges greater energy by heat transfer to the cold reservoir.
Answer: Attached below is the missing diagram
answer :
A) 1) Wr > WI, 2) Qc' > Qc
B) 1) QH' > QH, 2) Qc' > Qc
Explanation:
л = w / QH = 1 - Qc / QH and QH = w + Qc
A) each cycle receives same amount of energy by heat transfer
( Given that ; Л1 = 1/3 ЛR )
1) develops greater bet work
WR develops greater work ( i.e. Wr > WI )
2) discharges greater energy by heat transfer
Qc' > Qc
solution attached below
B) If Each cycle develops the same net work
1) Receives greater net energy by heat transfer from hot reservoir
QH' > QH ( solution is attached below )
2) discharges greater energy by heat transfer to the cold reservoir
Qc' > Qc
solution attached below
Select the correct statement(s) regarding IEEE 802.16 WiMAX BWA. a. WiMAX BWA describes both 4G Mobile WiMAX and fixed WiMax b. DSSS and CDMA are fundamental technologies used with WiMAX BWA c. OFDM is implemented to increase spectral efficiency and to improve noise performance d. all of the statements are correct
Answer:
d. all of the statements are correct.
Explanation:
WiMAX Broadband Wireless Access has the capacity to provide service up to 50 km for fixed stations. It has capacity of up to 15 km for mobile stations. WiMAX BWA describes both of 4G mobile WiMAX and fixed stations WiMAX. OFMD is used to increase spectral efficiency of WiMAX and to improve noise performance.
A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage of 120 V rms. The load draws a total of 10 kW at a power factor of 0.85 (lagging). Calculate the rms value of the phase currents and the magnitude of the per-phase load impedance. Draw a phasor diagram showing all tlme voltages and currents.
Answer:
Following are the solution to the given question:
Explanation:
Line voltage:
[tex]V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v[/tex]
Power supplied to the load:
[tex]P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi[/tex]
[tex]10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A[/tex]
Check wye-connection, for the phase current:
[tex]I_{ph}=I_L= 32.68\ A[/tex]
Therefore,
Phasor currents: [tex]32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A[/tex]
Magnitude of the per-phase load impedance:
[tex]Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega[/tex]
Phase angle:
[tex]\phi = \cos^{-1} \ (0.85) =31.79^{\circ}[/tex]
Please find the phasor diagram in the attached file.
Frame 4 questions about the challenges that astronauts have to face in space.
Answer:
Lack of oxygen
Decompression sickness
temperature variation
lack of gravity
Cosmic radiations hazards
motion sickness
Explanation:
When an astronaut travels in the space he is aware of the challenges he might face during his journey. There are various test and rehearsals of travelling before a final travel takes place. An astronaut goes through many challenges which includes lack of oxygen supply, decompression and motion sickness. There is no gravity in space so an astronaut will have to be aware of the difficulties he might face during his travel. An hour on earth is 7 years long in space, so an astronaut should have patience and be able to deal with time variation.
A resistivity meter is measured in
Your boss is given a new cylindrical aluminum part to be sand cast. The part is a
disk 50 cm in diameter and 20 cm thick to be cast of pure aluminum in a closed
mold system containing a gating system and riser volume that is 50% of the mold
cavity volume. Assume that a superheat of 100°C is used and that the aluminum
properties are as follows:
▪ Melting temperature of aluminum = 660°C
▪ Latent heat of fusion = 389.3 J/g density
▪ Density = 2.70 g/cm3
▪ Specific heat = 0.88 J/g-°C. Assume the specific heat has the same value for
solid and molten aluminum.
Q1:Compute the amount of heat (in MJ) that must be added to the metal to heat it to the pouring temperature, starting from a room temperature of 25°C.
Q2: Assume the same part in Q2 is being made with the same method. Based upon past experience when poured from the same temperature, the mold constant for sand casting an aluminum part was found to be 5 min/cm2. Determine the total solidification time (in minutes) for the part.
Q3:Assume that the sand mold downsprue is 25 cm long and the cross-sectional area at the base is 2.5 cm2. The downsprue feeds a horizontal runner leading into a mold cavity. Assume also that the part has the same dimensions as described in Question 22 and that the volumetric contraction for the cast metal in the mold is 4.3%. Finally, assume that using the current superheat, it takes 28 seconds before solidification begins. What is the mold fill time (in seconds)?
Q4: Will the mold fill before the start of freezing?
Q5: Assume that a part with unknown dimensions uses a permanent mold casting process with a mold constant of 1 min/cm2 and that the part solidifies in 60 min. Calculate the dimensions (in cm) of an effective riser assuming that the riser is a cylinder with a height/diameter ratio H/D = 1 and that the riser will take 10% longer than the casting to solidify. H = D = ?
What statement about the print() function is true?
print() has a variable number of parameters.
print() can have only one parameter.
print() can be used to obtain values from the keyboard.
print() does not automatically add a line break to the display.
Explanation:
print() has a variable number of parameters. this is the answer.
hope this helps you
have a nice day
write a verilog description of the following combinational circuit using concurrent statements. Each gate has a 5-ns delay, excluding the inverter, which has a 2-ns delay. (consider the below circuit is a full module)
Answer: Hello your question is incomplete attached below is the complete question
answer:
attached below
Explanation:
In this Verilog description we will refer to figure attached below
we will make some representation which are :
Represent outputs of the input AND gates = P
Represent outputs of the input NOR gates = Q
Inverter = R
attached below is the Verilog description
Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at 2208C with a mass flow rate of 1.2 kg/s. Refrigerant exits at 7 bar, 708C. Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in m3 /s, and (b) the power input to the compressor, in kW.'
Answer:
a)[tex]V_1=4.88m^2/s[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)[tex]P=-119.18kW[/tex]
Explanation:
From the question we are told that:
Steady State Saturated vapor [tex]T_1= -20C=>253k[/tex]
Mass Flow rate [tex]M=1.2kg/s[/tex]
Exit Pressure [tex]P_2=7bar[/tex]
Exit Temperature [tex]T_2=70C=>373k[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>P_1=1.399 bar[/tex]
Generally the equation for Volumetric Flow rate is mathematically given by
For Inlet
[tex]V_1=m\frac{RT_1}{P_1}[/tex]
[tex]V_1=m\frac{8314*253}{1.399*10^3}[/tex]
[tex]V_1=18.97m^2/s[/tex]
For outlet
[tex]V_2=m\frac{RT_2}{P_2}[/tex]
[tex]V_2=1.2*\frac{8314*343}{7*10^3}[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)
Generally the equation for Steady state mass and energy equation is mathematically given by
[tex]P=m(h_1-h_2)[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>h_1=24.76kJ/kg[/tex]
[tex]T_2= 70C =>h_2=124.08kJ/kg[/tex]
Therefore
[tex]P=1.2(12.76-124.08)[/tex]
[tex]P=-119.18kW[/tex]
Therefore
Power input into the compressor is
[tex]P=-119.18kW[/tex]
So I am going to do online school till I graduate and I have horrible internet. i only get about 3 quarters of each class I take so I miss most of it. WHAT DO I DO. my mom said she will never let me go back to a brick-and-mortar school.
In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2 and a Y value of 1.0. An in-service stress level of 200 MPa has been calculated. What is the length of a surface crack (in mm) that will lead to fracture
Answer:
[tex]l=24mm[/tex]
Explanation:
From the question we are told that:
Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]
Y value [tex]Y=1.0[/tex]
Stress level of[tex]\sigma =200 MPa[/tex]
Generally the equation for length of a surface crack is mathematically given by
[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]
[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]
[tex]l=0.024m[/tex]
Therefore
in mm
[tex]l=24mm[/tex]
Why water parameters of Buriganga river vary between wet and dry seasons?
Explain.
A brittle failure has extensive plastic deformation in the vicinity of the advancing crack. This process proceeds relatively slow (stable) and the crack resists any further extension unless there is an increase in the applied stress
a. True
b. False
Answer:
False ( b )
Explanation:
In a brittle failure the cracks spreads rapidly without a significant deformation, and the cracks are very unstable with the cracks extending without an increase in the amount of applied stress.
Therefore the above description in the question is false.
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heated uniformly to115 oC, determine the length and the diameter of this rod to the nearest micron at the new temperature if the linear coefficient of thermal expansion of steel is 12.5 x 10-6 m/m/oC. What is the stress on the rod at 115oC.
Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fluid flow through the 3.0-inch section will be what factor times that through the 1.0-inch section
Answer:
[tex](\frac{r_1}{r_2})^2=\frac{1}{9}[/tex]
Explanation:
From the question we are told that:
Diameter 1 [tex]d_1=1.0[/tex]
Diameter 2 [tex]d_2=3.0[/tex]
Generally the equation for Radius is mathematically given by
At Diameter 1
[tex]r_{1}=\frac{1}{2} inch[/tex]
At Diameter 2
[tex]r_{2}=\frac{3}{2} inch[/tex]
Generally the equation for continuity is mathematically given by
[tex]A_1V_1=A_2V_2[/tex]
Therefore
[tex](\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2[/tex]
[tex](\frac{r_1}{r_2})^2=\frac{1}{9}[/tex]
The value of universal gas constant is same for all gases?
a) yes
b)No
Answer:
The answer of these questions is
Explanation:
b) NO
For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.
a. G(s)= 1/s(s+2)(s+4)
b. G(s)= (s+5)/(s+2)(s+4)
c. G(s)= (s+3)(s+5)/s(s+2)(s+4)
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
state two disadvantages and two advantages of a simple manometer.
Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ cấu ở trạng thái cân bằng. Tại thời điểm khâu 1 ở vị trí υ1 = 900 hãy tính áp lực tại các khớp động tại B, C và A.
Calculate the number of 12 V batteries (capacity 120 Ah) needed to run a 3 kW DC motor that operates in 240 V. How many hours the motor will run with 20 of such batteries connected in series?
Answer:
20 batteries9.6 hoursExplanation:
To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...
240/12 = 20 . . . batteries needed
__
The current draw will be ...
(3000 W)/(240 V) = 12.5 A
Then the time available from the battery stack is ...
(120 Ah)/(12.5 A) = 9.6 h
The motor can run 9.6 hours from the series connection.
Show that a short-circuited losalesa transmission line looks at its input end as an inductor or a capacitor. This property is used in integrated microwave circuits to simulate a capacitor or an traductor.
Answer:
A section of a lossless transmission line short circuited acts a circuit reactive element ( i.e. inductor or capacitor ) and to achieve the desired reactance, the length of the transmission line has to be properly chosen.
attached below is the related diagram
Explanation:
Input impedance of a short-circuited lossless transmission line can be expressed as :
attached below is the reaming part of the solution
A section of a lossless transmission line short circuited acts a circuit reactive element ( i.e. inductor or capacitor ) and to achieve the desired reactance, the length of the transmission line has to be properly chosen. which is used in stub matching.
a basketball player pushes down with a force of 50N on a basketball that is indlated to a gage pressure of 8.0x10^4 Pa. What is the diameter of comtact between the ball nad the floor
Answer:
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Explanation:
The basketball experiments a normal stress ([tex]\sigma[/tex]), in pascals, due to normal force from the floor ([tex]N[/tex]). By definition of normal stress, we have the following equation:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex] (1)
Where [tex]D[/tex] is the diameter of the contact area between the ball and the floor, in meters.
Please notice that magnitude of the normal force equals the magnitude of external force given by the basketball player and weight is negligible in comparison with normal and external forces.
If we know that [tex]N = 50\,N[/tex] and [tex]\sigma = 8.0\times 10^{4}\,Pa[/tex], then the diameter of the contact area is:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex]
[tex]D^{2} = \frac{4\cdot N}{\pi\cdot \sigma}[/tex]
[tex]D = 2\cdot \sqrt{\frac{N}{\pi\cdot \sigma} }[/tex]
[tex]D = 2\cdot \sqrt{\frac{50\,N}{\pi\cdot (8\times 10^{4}\,Pa)} }[/tex]
[tex]D\approx 0.028\,m[/tex] [tex](28\,mm)[/tex]
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
