Answer:
some of the professions and human resource related to engineering are:
Aerospace engineerAgricultural engineer Electrical engineer computer engineerproject Manger construction site engineer/supervisorAerospace engineering involves the study, design and development of spacecrafts using Core science principles.
Electrical engineering involves the study and application of core science principles especially physics and mathematics into providing Electrical related solutions
Explanation:
Engineering is a major branch of applied science. In general Engineering is concerned with the design and building of engines ( i.e. application of scientific/science facts )
some of the professions and human resource related to engineering are:
Aerospace engineerAgricultural engineer Electrical engineer computer engineerproject Manger construction site engineer/supervisorAerospace engineering involves the study, design and development of spacecrafts using Core science principles.
Electrical engineering involves the study and application of core science principles especially physics and mathematics into providing Electrical related solutions
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
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
What is the formula for resultant force
Answer:
F = 3 N + 2 N = 5 N
Explanation:
Resultant force F = 3 N + 2 N = 5 N to the right. The resultant force is 5 N to the right. Two forces that act in opposite directions produce a resultant force that is smaller than either individual force. To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force.
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
A(n) ____ combines two planetary gearsets to provide more gear ratio possibilities. A)compound planetary gearset B)orifice C)detent D)lock-up torque converter
Answer:
The answer is A. Compound Planetary Gearset.
Explanation:
The Compound Planetary Gear block represents a planetary gear train with composite planet gears. Each composite planet gear is a pair of rigidly connected and longitudinally arranged gears of different radii. One of the two gears engages the centrally located sun gear while the other engages the outer ring gear.
Compound planetary gear sets have at least two planet gears attached in line to the same shaft, rotating and orbiting at the same speed while meshing with different gears. Compounded planets can have different tooth numbers, as can the gears they mesh with.
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].
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 route for a proposed 8-m-wide highway crosses a region with a 4-m-thick saturated, soft, normally consolidated clay (CH) above impermeable rock. Groundwater level is 1 m below the surface. The geotechnical data available during the preliminary design stage consist of Atterberg limits (LL 5 68% and PL 5 32%) and the natural water content (w 5 56%). Based on experience, the geotechnical engineer estimated the coefficient of consolidation at 8 m2 per year. To limit settlement, a 4-m-high embankment will be constructed as a surcharge from fill of unit weight 16 kN/m3.
(a) Estimate the compression and recompression indices.
(b) Estimate the total primary consolidation settlement under the center of the embankment.
(c) Plot a time–settlement curve under the center of the embankment.
(d) How many years will it take for 50% consolidation to occur?
(e) Explain how you would speed up the consolidation.
(f) Estimate the rebound (heave) when the surcharge is removed.
Sketch a settlement profile along the base of the embankment. Would the settlement be uniform along the base? Explain your answer.A route for a proposed 8-m-wide highway crosses a
These waveforms are applied to a gated D latch, which is initially RESET. Which of the areas identified on the Q waveform is incorrect?
Question Completion with Options:
A) Area a B) Area b C) Area c D) Area d
Answer:
The incorrect waveform identified on the Q waveform is the:
C) Area c.
Explanation:
Area c is the incorrect waveform because its output is not correct. The Q waveform indicates that the electrical forces project toward the negative pole of the lead axis. A gated D latch is a flip flop latch with an additional control input, which determines when to change the state of the circuit. Most times, this control unit is a clock input or an enable input.
A resistivity meter is measured in
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
Assignment Using Perman's equation, estimate the potential evapotranspiration for the month of August in locality with the following data
latitude 20 degrees north
Elevation 200m
• Mean monthly temperature 20 degreeso
• Mean relative humidity: 75%
• Mean sunshine hour= 9hrs
Wind at 2 m height equal 85 km per day
nature of surface cover =green grass
Answer:
what do you need help with
Explanation:
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.
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.
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
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
(a) What are the usual causes of mechanical failure in the component or system? ( b ) What are the general types of mechanical failure ?
Answer:
Some types of mechanical failure mechanisms are: excessive deflection, buckling, ductile fracture, brittle fracture, impact, creep, relaxation, thermal shock, wear, corrosion, stress corrosion cracking, and various types of fatigue. ... Cascading failures, for example, are particularly complex failure causes.
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.
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)
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/tex]
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
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]
Cho thanh có tiết diện thay đổi chịu tải trọng dọc trục (hình 1).
Biết d1 = 5 cm, d2 = 8 cm, a= 15 cm, b=10cm, P1 =400kN, P2 =200kN, E= 2.104 kN/cm2.
a) Vẽ biểu đồ lực dọc.
b) Kiểm tra bền của thanh AC, [ϭ] =10 (kN/cm2).
c) Xác định chuyển vị theo phương dọc trục của tâm tiết diện C
Answer:
saay in English language
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 = ?
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.
What is the built-in pollution control system in an incinerator called
Explanation:
hbyndbnn☝️
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)
You are working for a company that creates special magnetic environments. Your new supervisor has come from the financial side of the organization rather than the technical side. He has promised a client that the company can provide a device that will create a magnetic field inside a cylindrical chamber that is directed along the cylinder axis at all points in the chamber and increases in the axial direction as the square of the value of y, where y is in the axial direction and y = 0 is at the bottom end of the cylinder. Prepare a calculation to show that the field requested by your supervisor and promised to a client is impossible.
Answer:
Following are the responses to the given question:
Explanation:
When we take the entire cylinder as a surface that is:
[tex]B=B.y^2\ j\\\\[/tex]
for the magnetic filed existance
[tex]\bigtriangledown . \underset{b}{\rightarrow}=0[/tex]
[tex]\therefore[/tex]
the flub by this cyclinder is zero implies there theaming flubs = outgoing flubs
[tex]\bigtriangledown . \underset{b}{\rightarrow}\\\\=\frac{\delta }{\delta x} B_x+\frac{\delta }{\delta y } B_y+ \frac{\delta }{\delta z} B_z\\\\=\frac{\delta }{\delta x} B_0+\frac{\delta }{\delta y } B_{y^2}+ 0\\\\=\frac{\delta }{\delta x} B_0\ {y^2}=2 B_0\ y\\\\So,\\\\\bigtriangledown . \underset{b}{\rightarrow}\neq 0[/tex]
that's why it is impossible field.
