Answer:
- weather refers to short-term changes in the atmosphere
- climate describes what the weather is like over a long period of time in one area.
A concrete batch plant has a batching chamber capacity of 12.5 cy and an average batching cycle time of 3 min. Plant efficiency is 84%. What is the estimated batching production in cy/hr?
Answer:
The estimated batching production is 210 cy/hr
Explanation:
Given;
capacity of the batching chamber of the plant, Q = 12.5 cy
average batching cycle time, t = 3 min
plant efficiency, n = 84 %
The estimated batching production is calculated as efficiency of the machine multiplied by the production rate in hour. This is given by the solution below;
Batching production = (nQ )/ t
[tex]Batching \ production = 0.84(\frac{12.5 \ cy}{3 \ min} *\frac{60 \ min}{1 \ hr})\\\\ Batching \ production = 210 \ cy/hr[/tex]
Therefore, the estimated batching production is 210 cy/hr
Estimate the rotor inertia assuming that the rotor is a cylinder of radius 8.98 mm, and length 25 mm, with a material of 100% copper. Explain why the rotor inertia may differ from these assumptions?
Answer:
The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.
The rotor inertia may differ from these assumption due to differences in the shape of cross section.
Explanation:
We assume that rotor can be represented as a solid cylinder of radius [tex]r[/tex], length [tex]l[/tex], made of cooper ([tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex]) and whose axis of rotation passes through its center of mass and is parallel to its cross section. By definition of Moment of Inertia and Theorem of Parallel Axes, the moment of inertia of the rotot is:
[tex]I = \frac{1}{4}\cdot \rho \cdot \left(\frac{\pi}{4} \right) \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex]
[tex]I = \frac{\pi}{16}\cdot \rho \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of copper, measured in kilograms per cubic meter.
[tex]R[/tex] - Radius of the rotor, measured in meters.
[tex]L[/tex] - Length of the rotor, measured in meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
If we know that [tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex], [tex]L = 25\times 10^{-3}\,m[/tex] and [tex]R = 8.98\times 10^{-3}\,m[/tex], the estimated moment of inertia of the rotor is:
[tex]I = \frac{\pi}{16}\cdot \left(8960\,\frac{kg}{m^{3}} \right)\cdot (8.98\times 10^{-3}\,m)^{3}\cdot [3\cdot (8.98\times 10^{-3}\,m)^{2}+(25\times 10^{-3}\,m)^{2}][/tex]
[tex]I \approx 1.105\times 10^{-6}\,kg\cdot m^{2}[/tex]
The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.
From D'Alembert's Formula we know that net force of rigid bodies experimenting rotation equals the product of moment of inertia and angular acceleration. In this case, the purpose is minimizing moment of inertia and it is done by modifying the shape of the cross section so that rotor could be aerodynamically more efficient.
Facts about cellphones
Answer:
Your mobile phone has more computing power than the computers used for the Apollo 11 moon landing.
Mobile phones have to “work harder” to get a signal if you are in a moving vehicle.
The first mobile phone was made in 1973.
The first mobile phones that went on sale in 1983 cost nearly $4,000 each.
In 2012 Apple sold 340,000 phones per day.
4 out of 10 Brits admit to snooping on their partners phone.
Out of the 53% of snoopers that found incriminating evidence on their partner’s phone, 5% went on to terminate their relationship.
Waterproof mobile phones came to market because Japanese youngsters like to use them in the shower.
Apparently mobile phones have 18 times more bacteria on them than toilet handles!
Phubbing describes the act of snubbing someone by using your mobile phone in their company.
In 2015 more people died from taking selfies than shark attacks.
Teenagers that use a phone more than 2 hours a day increase their risk of depression and anxiety.
Nomobophobia is severe anxiety caused by the thought or act of losing your phone or running out of battery.
Explanation:
The critical resolved shear stress for a metal is 36 MPa. Determine the maximum possible yield strength (in MPa) for a single crystal of this metal that is pulled in tension.
Answer:
72 MPa
Explanation:
Critical resolved shear stress = 36 MPa
calculate the maximum possible yield strength for a single crystal of the metal
first we have to express the critical resolved shear stress as
бy = 2т[tex]_{critical}[/tex]
given that the minimum stress needed to introduce yielding curve will occur at Ψ = λ = 45
where : бy = yielding stress
t[tex]_{critical}[/tex] = 36MPa
therefore the maximum possible yield strength
= (2) * (36)
= 72 MPa
It is desired to enrich the partial pressure of hydrogen in a hydrogen–nitrogen gas mixture for which the partial pressures of both gases are 0.1013 MPa (1 atm). It has been proposed to accomplish this by passing both gases through a thin sheet of some metal at an elevated temperature; in as much as hydrogen diffuses through the plate at a higher rate than does nitrogen, the partial pressure of hydrogen will be higher on the exit side of the sheet. The design calls for partial pressures of 0.051 MPa (0.5 atm) and 0.01013 MPa (0.1 atm), respectively, for hydrogen and nitrogen. The concentrations of hydrogen and nitrogen (CHC
H and CNC N , in mol/m3mol/m
3 ) in this metal are functions of gas partial pressures (pH2 and pN2p
H 2 and p N , in MPa) and absolute temperature and are given by the following expressions:
CH=2.5×103√pH2exp(−27,800J/mol/RT)
CN=2.75×103√pN2exp(−37,600J/mol/RT )
Furthermore, the diffusion coefficients for the diffusion of these gases in this metal are functions of the absolute temperature, as follows:
DH(m2/s)=1.4×10−7exp(−13,400J/mol/RT)
DN(m2/s)=3.0×10−7exp(−76,150J/mol/RT)
Is it possible to purify hydrogen gas in this manner? If so, specify a temperature at which the process may be carried out, and also the thickness of metal sheet that would be required. If this procedure is not possible, then state the reason(s) why.
Answer:
T = 3460 K
Explanation:
See attachment for calculation.
Since the temperature we have is above the melting point of the metal, then we can conclude that it is too high for the diffusion process to be possible.
Recently, due to rapid urbanization and mechanization residents of a city are suffering from harmful effects of the ultra violet rays. Depletion of which layer is likely to have led to this situation
A hypothetical metal has the simple cubic (SC) crystal structure. If its atomic weight is 70.4 g/mol and the atomic radius is 0.126 nm. Compute its density (in g/cm3)
Answer:
[tex]7304g/cm^3[/tex]
Explanation:
Density is the mass per unit volume, it is usually measured in kg/m³.
The density of a crystal structure is given by:
[tex]\rho=\frac{mass\ of\ atoms\ in\ unit\ cell}{volume\ unit\ cell} =\frac{nA}{V_cN_A} \\\\\rho=density, n= number\ of\ atoms/unit=1\ atom/unit\ cell,A=atomic\ weight=70.4\ g/mol,N_A=avogadro\ constant=6.023*10^{23}atoms/mol,V=volume=2R^3=[2(0.126*10^{-8})]^3cm\\\\\rho=\frac{1*70.4}{[2(0.126*10^{-8})]^3*6.023*10^{23}} =7304g/cm^3[/tex]
Answer:
7.307g/cm^3
Explanation:
The other guy is mostly right but his conversion factor was off. It should be 1.26x10^-8 on the bottom
gasoline has a comparatively high btu per galloon rating around?
Answer:
116,090 Btus
Explanation:
Your company is planning to build a pipeline to transport gasoline from the refinery to a field of storage tanks. The parameters for the prototype system are a pipe diameter of 1 m, with a flow velocity of 0.5 m/s at 25°C. The model system will use water at STP with a geometric scaling factor of 1 : 20. What fluid velocity is required in the model system to guarantee kinematic similarity in the form of equal Reynolds numbers?
Answer:
The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.
Explanation:
The Reynolds number ([tex]Re_{D}[/tex]) is a dimensionless criterion use for flow regime of fluids, which is defined as:
[tex]Re_{D} = \frac{\rho \cdot v\cdot D}{\mu}[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density, measured in kilograms per cubic meter.
[tex]\mu[/tex] - Dynamic viscosity, measured in kilograms per meter-second.
[tex]v[/tex] - Average flow velocity, measured in meters per second.
[tex]D[/tex] - Pipe diameter, measured in meters.
We need to find the equivalent velocity of water used in the prototype system. In this case, we assume that [tex]Re_{D,gas} = Re_{D,w}[/tex]. That is:
[tex]\frac{\rho_{w}\cdot v_{w}\cdot D_{w}}{\mu_{w}} = \frac{\rho_{gas}\cdot v_{gas}\cdot D_{gas}}{\mu_{gas}}[/tex] (Eq. 2)
Where subindex [tex]w[/tex] is used for water and [tex]gas[/tex] for gasoline.
If we know that [tex]\rho_{gas} = 690\,\frac{kg}{m^{2}}[/tex], [tex]\mu_{gas} = 0.006\,\frac{kg}{m\cdot s}[/tex], [tex]v_{gas} = 0.5\,\frac{m}{s}[/tex], [tex]D_{gas} = 1\,m[/tex], [tex]\rho_{w} = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\mu_{w} = 0.0018\,\frac{kg}{m\cdot s}[/tex] and [tex]D_{w} = 0.05\,m[/tex], then we get the following formula:
[tex]57500 = 27777.778\cdot v_{w}[/tex]
The fluid velocity for the prototype system is:
[tex]v_{w} = 2.07\,\frac{m}{y}[/tex]
The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.
What is the meaning of *binuhat lakas nang loob na ibinaon
Answer:
carried courage buried
Explanation:
its Filipino
Nicole designs the hardware configuration of workstations that will be deployed to a newly formed company. She sets up the networking capabilities and policies that will govern the workstations when connected to the company network. What is her role in her company?
Answer:
a Network Engineer (architect)
Explanation:
Indeed, as a Network Engineer, we would expect Nicole to be in charge of planning, setting up, and managing the software and hardware components of the computer networks so that they function as intended.
A Network Engineer therefore would be responsible for setting up the networking capabilities and policies that will govern the workstations when connected to the company network.
A commercial jet is flying at a standard altitude of 35,000 ft with a velocity of 550 mph: (a) what is the Mach number? (b) should the flow be treated as incompressible, why or why not?
Answer:
Mach number = 0.68168
The flow should be treated as compressible.
Explanation:
Given that:
The altitude of a commercial jet = 35000
The properties of air at that given altitude are as follows:
Pressure = 24.577 kPa
Temperature T = 50.78176° C
Temperature T = ( 50.78176 + 273 )K = 328.78176 K
[tex]\varphi = 0.38428 \ kg/m^3[/tex]
The velocity is also given as: 550 mph = 245.872 m/s
Therefore, the sonic velocity is firstly determined by using the formula:
[tex]a = \sqrt{ \vartheta \times R \times T\[/tex]
[tex]a = \sqrt{1.4 \times 287 \times 323.78176[/tex]
[tex]a = \sqrt{130095.5112[/tex]
a = 360.68755 m/s
Then, we can calculate the Mach number by using the expression:
[tex]{Mach \ number = \dfrac{V}{a}}[/tex]
[tex]Mach \ number = \dfrac{245.872}{360.68755}[/tex]
Mach number = 0.68168
b) Ideally, all flows are compressible because the Mach number is greater than 0.3, suppose the Mach number is lesser than 0.3, then it is incompressible.
How natural gas works and operates?
PLEASE HELP!!!
A car makes a hissing noise each time the A/C system and engine are turned off. Technician A says that the noise is caused by a refrigerant leak. Technician B says that the noise is caused by equalization of system pressures. Who is correct?
Answer:
It's equalization of the system
Explanation: If there was a leak it would likely leak all the time even if the car was turned off. Plus, a system leak bad enough to hear would drain the system of refrigerant very quickly and would no longer cool.
Use the following clues to help fill in the chart on the next page. Put an X in the spaces that are INCORRECT and Highlight the
CORRECT
1 The wizard with the lavender wand is in Ravenel or Sparrowan, and earned 50 or 60 points
2. Gorgonscale earned 10 points less than Sparrowman
3 Lynn scored 20 points less than the wizard with the incense wand.
4. Timmy scored 70 or 80 points. He is in Gorgonscale or Hydraden
5. Among Bennie and the wizard from Sparrowan, one earned 70 points and the other has the lavender wand.
6. The mandragore wand belongs to Edward or to the House of Hydraden
7 Ravenel didn't earn 60 points and Edward is not among it's wizards
8. Bennie scored 10 points more than Edward
9. The wizard with the mandragore wand didn't earn 70 points.
inspections may be_____ or limited to a specific area such as electrical or plumbing
A. Metering
B. General
I need help on this ASAP PLZ?
Answer:
Explanation:
7. False
8. True
9. True
10.True
11. True
(TCO 7) Find the resolution of the Fourier transform if a signal is sampled at 16,384 samples/second and we collect a total of 8,192 data points and then apply the FFT algorithm to them.
Answer:
Resolution of Fourier transform = 2Hz
Explanation:
Given:
Sample signals = 16,384 samples/second
Number of data points = 8,192 oints
Find:
Resolution of Fourier transform
Computation:
Resolution of Fourier transform = fs / N
Resolution of Fourier transform = 16,384 / 8,192
Resolution of Fourier transform = 2Hz
Should transistors used in switching circuits be biased in the active region? Why or why not?
Answer:
no
Explanation:
No. More power is dissipated when the transistor is in its active region. In general, transistors in switching circuits are biased either "on" or "off". Time spent in the active region is minimized.
_____
On the other hand, speed can be enhanced if the transistors are active. So, it's a speed/power trade-off. Usually power is of more interest, particularly when there are millions of switching circuits. However, in certain applications, speed may be the priority, so the transistor will be biased in its active region.
To determine if a product or substance being used is hazardous, consult:__________.
a. A dictionary
b. An MSDS
c. SAE standards
d. EPA guidelines
Answer:
Option B: An MSDS
Explanation:
A dictionary is used to check up the meaning of general words and not for checking if a substance being used is hazardous. Option A is wrong.
MSDS means "Material Safety Data Sheet" and it contains documents with information that relates to occupational health & safety for checking various substances and products. Thus, option B is correct.
SAE stands for Society of Automotive Engineering and their standards pertain to mainly Automobiles. Thus option C is wrong.
EPA guidelines are mainly for checking facility and environmental health and safety compliance. Thus, option D is wrong.
Realiza las siguientes conversiones.
4 Hm2 a Dm2=_______________
21345 Cm2 a M2=_____________
0,592 Km2 a M2=______________
0,102 M2 a Cm2=______________
23911 Km2 a Hm2=_____________
Answer:
a) 4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.
b) 21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.
c) 0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.
d) 0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.
e) 23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.
Explanation:
a) 4 hectómetros cuadrados a decámetros cuadrados:
Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un hectómetro cuadrado equivale a 100 decámetros cuadradps. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:
[tex]x = 4\,Hm^{2}\times\frac{100\,Dm^{2}}{1\,Hm^{2}}[/tex]
[tex]x = 400\,Dm^{2}[/tex]
4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.
b) 21345 centímetros cuadrados a metros cuadrados:
Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:
[tex]x = 21345\,cm^{2}\times \frac{1\,m^{2}}{10000\,cm^{2}}[/tex]
[tex]x = 2,135\,m^{2}[/tex]
21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.
c) 0,592 kilómetros cuadrados a metros cuadrados:
Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 1000000 metros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:
[tex]x = 0,592\,km^{2}\times \frac{1000000\,m^{2}}{1\,km^{2}}[/tex]
[tex]x = 592000\,m^{2}[/tex]
0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.
d) 0,102 metros cuadrados a centímetros cuadrados:
Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:
[tex]x = 0,102\,m^{2}\times \frac{10000\,cm^{2}}{1\,m^{2}}[/tex]
[tex]x = 1020\,cm^{2}[/tex]
0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.
e) 23911 kilómetros cuadrados a hectómetros cuadrados:
Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 100 hectómetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:
[tex]x = 23911\,km^{2}\times \frac{100\,Hm^{2}}{1\,km^{2}}[/tex]
[tex]x = 2391100\,Hm^{2}[/tex]
23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.
A 30-mm-diameter shaft, made of AISI 1018 HR steel, transmits 10 kW of power while rotating at 200 rev/min. Assume any bending moments present in the shaft to be negligibly small compared to the torque. Determine the static factor of safety based on:a) The maximum-shear-stress failure theory.b) The distortion-energy failure theory.
Answer:
a) According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.
b) According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.
Explanation:
First, we need to determine the torque experimented by the shaft ([tex]T[/tex]), measured in kilonewton-meters, whose formula is described:
[tex]T = \frac{\dot W}{\omega}[/tex] (Eq. 1)
Where:
[tex]\dot W[/tex] - Power, measured in kilowatts.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
If we know that [tex]\dot W = 10\,kW[/tex] and [tex]\omega = 20.944\,\frac{rad}{s}[/tex], then the torque experimented by the shaft:
[tex]T = \frac{10\,kW}{20.944\,\frac{rad}{s} }[/tex]
[tex]T =0.478\,kN\cdot m[/tex]
Let consider that shaft has a circular form, such that shear stress is determined by the following formula:
[tex]\tau = \frac{16\cdot T}{\pi\cdot D^{3}}[/tex] (Eq. 2)
Where:
[tex]D[/tex] - Diameter of the shaft, measured in meters.
[tex]\tau[/tex] - Torsional shear stress, measured in kilopascals.
If we know that [tex]D = 0.03\,m[/tex] and [tex]T =0.478\,kN\cdot m[/tex], the torsional shear stress is:
[tex]\tau = \frac{16\cdot (0.478\,kN\cdot m)}{\pi\cdot (0.03\,m)^{3}}[/tex]
[tex]\tau \approx 90164.223\,kPa[/tex]
a) According to the maximum-shear-stress failure theory, we get that maximum shear stress limit is:
[tex]S_{ys} = 0.5\cdot S_{ut}[/tex] (Eq. 3)
Where:
[tex]S_{ys}[/tex] - Ultimate shear stress, measured in kilopascals.
[tex]S_{ut}[/tex] - Ultimate tensile stress, measured in kilopascals.
If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:
[tex]S_{ys} = 0.5\cdot (440\times 10^{3}\,kPa)[/tex]
[tex]S_{ys} = 220\times 10^{3}\,kPa[/tex]
Lastly, the static factor of safety of the shaft ([tex]n[/tex]), dimensionless, is:
[tex]n = \frac{S_{ys}}{\tau}[/tex] (Eq. 4)
If we know that [tex]S_{ys} = 220\times 10^{3}\,kPa[/tex] and [tex]\tau \approx 90164.223\,kPa[/tex], the static factor of safety of the shaft is:
[tex]n = \frac{220\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]
[tex]n = 2.440[/tex]
According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.
b) According to the distortion-energy failure theory, we get that maximum shear stress limit is:
[tex]S_{ys} = 0.577\cdot S_{ut}[/tex] (Eq. 5)
If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:
[tex]S_{ys} = 0.577\cdot (440\times 10^{3}\,kPa)[/tex]
[tex]S_{ys} = 253.88\times 10^{3}\,kPa[/tex]
Lastly, the static factor of safety of the shaft is:
[tex]n = \frac{253.88\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]
[tex]n = 2.816[/tex]
According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.
Two vehicles collided head on while traveling on a curve tangent with a 3% grade. Vehicle V1 skidded 195 feet downhill before colliding with vehicle V2. Vehicle V2 skidded 130 feet. The police report estimates that the speed of both vehicles at impact was 25 mph, based on vehicle deformation. Assume a coefficient of friction of 0.48. What is V1 speed at the beginning of the skid?
Answer:
Speed V1 at the beginning of the skid = 57.075 mph
Explanation:
Calculate the V1 speed at the beginning of the skid
V1 ( speed before brake was applied ) = ?
distance travelled by Vehicle 1 = 195 ft
curve tangent for vehicle 1 = - 3% = - 0.03
coefficient of friction = 0.48
speed at Impact = 25 mph
To determine the V1 speed at the beginning of the skid we have to apply AASTHO
d1 = [tex]( \frac{VA^{2} - VA^{12} }{30(F+N)} )[/tex]
195 = [tex](\frac{VA^2- 25^2}{30(0.48-0.03)} )[/tex]
Hence : VA = 57.075 mph