Answer:
How to stop toilets
Explanation:
I think
Hope this helps
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
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)
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.
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
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.
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.
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
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
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.
Future solution for air pollution in new zealand
Answer:
New Zealand may use some of these solutions to prevent air pollution
Explanation:
Using public transports.
Recycle and Reuse
No to plastic bags
Reduction of forest fires and smoking
Use of fans instead of Air Conditioner
Use filters for chimneys
Avoid usage of crackers
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
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.
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
1. Is the team developing confidence that the solution space has been fully explored?
2. Are there alternative diagrams and alternative ways to decompose the problem?
3. Have external sources been thoroughly pursued, and everyone’s ideas been accepted and integrated in the process?
4. All of the above
Answer:
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
4. All of the above
Explanation:
The team must explore its solution space, including some external sources. Then, it must integrate its findings with the ideas of team members, ensuring the consideration of all possible ways to decompose the problem. This is because employing a structured process to concept generation enables the team to come up with creative solutions to design concepts.
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
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]
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ____________ of the pipe.
Answer:
Inlet
Explanation:
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.
This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.
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.
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:
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]
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
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.
Explain the LWD process why is it important in drilling operations?
Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.
Explanation:
pls mark brainliest
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
what is the term RF exiciter?
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.
What is the built-in pollution control system in an incinerator called
Explanation:
hbyndbnn☝️
Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend auaching fins both inside and outside the tubes?
Answer:
Fins should be attached outside the tube Fins can be attached on both sides when convection coefficient of air inside the tube is equal to the convection coefficient of atmospheric air outside the tubeExplanation:
The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.
The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ), BUT
When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )
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
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.
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]
