Answer:
THE thing is 435
Explanation:
Some of the trade-offs that must be made while transitioning from an individually tailored solution to such an enterprise-level solution are about as continues to follow.
Enterprise-level analytics Different individuals must be given some amount of control over their data in corporate-level systems.To counteract opposition from individuals who seem to be comfortable working utilizing customized statistics, training needs to be provided so that they can utilize cutting-edge large enterprise technologies.Adequate orientation training, as well as perks, should have been provided so that employees may become acquainted using the changing approach.
As a result, the aforementioned response seems to be correct.
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Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
Hope this helps...........
Find the complex power, average power, and reactive power for the following:
v(t) = 339.4 sin (377t + 90°) V, i(t) = 5.657 sin (377t + 60°) A
Answer:
Explanation:
Given that:
v(t) = 339.4 sin(377t + 90°) V
i(t) = 5.657 sin (377t + 60°) A
v = 339.4 ∠ 90° [tex]v_m \angle\phi_1[/tex]
i = 5.657∠60° [tex]I_m \angle \phi_2[/tex]
The phase difference [tex]\phi = 90 -60[/tex] = 30
The average power [tex]P_{avg}[/tex] can be expressed as:
[tex]P_{avg} = \dfrac{v_m}{\sqrt{2}}\dfrac{I_m}{\sqrt{2}} \times cos (30)[/tex]
[tex]P_{avg} = \dfrac{339.4}{\sqrt{2}}*\dfrac{5.657}{\sqrt{2}} \times cos (30)[/tex]
[tex]\mathbf{P_{avg} = 831.38 \ watts}[/tex]
The reactive power Q is as follow;
[tex]Q = \dfrac{v_m}{\sqrt{2}} * \dfrac{I_m}{\sqrt{2}} \ sin \phi\\[/tex]
[tex]Q = \dfrac{339.4}{\sqrt{2}}*\dfrac{5.657}{\sqrt{2}} \times sin (30)[/tex]
Q = 479.99 VAR
The complex power S = P + jQ
The complex power S = 831.38 W + j479.99 VAR
A cylindrical specimen of a brass alloy having a length of 104 mm (4.094 in.) must elongate only 5.20 mm (0.2047 in.) when a tensile load of 101000 N (22710 lbf) is applied. Under these circumstances what must be the radius of the specimen? Consider this brass alloy to have the stress–strain behavior
Answer:
The radius of the specimen is assumed to be 9.724 mm
Explanation:
Given that:
For a cylindrical specimen of a brass alloy;
The length = 104 mm, Elongation = 5.20 mm and the tensile load = 101000 N
Let's first determine the radius of the cylindrical brass alloy from the knowledge of the cross-sectional area of a cylinder.
[tex]A_0 = \pi r ^2[/tex]
[tex]r = \sqrt{\dfrac{A_o}{\pi}}[/tex]
[tex]r = \sqrt{\dfrac{\bigg (\dfrac{F}{\sigma} \bigg )}{\pi}}[/tex]
[tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]
To estimate the tensile stress:
We need to first determine the strain relating to elongation at 5.20 mm
[tex]Strain \ \ \varepsilon= \dfrac{\Delta l}{l_o}[/tex]
[tex]Strain \ \ \varepsilon= \dfrac{5.20}{104}[/tex]
Strain ε = 0.05
Using the stress-strain plot; let assume that under the circumstances; [tex]\sigma[/tex] = 340 MPa for stress corresponding to 0.05 strain
Thus;
The cylindrical brass alloy radius [tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]
[tex]r =\sqrt{ \dfrac{101000}{(340\times 10^{6})\pi}[/tex]
r = 0.009724 m
r = 9.724 mm
1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?
Answer:
a) up
b) down
Explanation:
When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).
a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.
__
b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.
A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?
Answer:
required feedback resistance ( R2 ) = 100 k Ω
Explanation:
Given data :
Voltage gain = 100
input resistance ( R1 ) = 1 k ohms
calculate feedback resistance required
voltage gain of differential amplifier
[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]
= Voltage gain = R2/R1
= 100 = R2/1
hence required feedback resistance ( R2 ) = 100 k Ω
Water flows at 10 m3/s in a 5-m-wide channel. What is the height of a suppressed rectangular (sharp-crested) weir that will cause the depth of flow in the channel to be 2 m
Answer:
Hw = 1.01 meters
Explanation:
Given data:
flow rate = 10 m^3
depth of flow in channel = 2 m
Determine the height of a suppressed rectangular weir ( Hw ) using the following expressions
expression for the elevation of of water surface above crest of weir
H = 2 - [tex]H_{w}[/tex] ------ ( 1 )
expression for the height of the weir ( Hw )
Hw = 2 - [tex]( \frac{Q}{C_{w} b}) ^{\frac{3}{2} }[/tex] ---------- ( 2 )
expression for the weir coefficient ( Cw )
Cw = [tex]\frac{2}{3} C_{d} \sqrt{2g}[/tex] -------------- ( 3 )
expression for the coefficient of discharge ( Cd )
Cd = 0.611 + 0.075 [tex]\frac{H}{Hw}[/tex] ---------- ( 4 )
Finally to determine the value of Hw we apply the trial and error method
in the trial and error method the value of LHS = RHS for the number chosen to be true
How can technology interfere with good study habits?
Answer:
the third anwser is right on edgeinuity
Explanation:
Answer:
Technology can pull students’ focus away from their tasks.
Explanation:
Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate of 1.3 m3/min. The work input to the control volume is 240 kJ per kg of air flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW.
Answer:
Heat transfer = 2.617 Kw
Explanation:
Given:
T1 = 300 k
T2 = 440 k
h1 = 300.19 KJ/kg
h2 = 441.61 KJ/kg
Density = 1.225 kg/m²
Find:
Mass flow rate = 1.225 x [1.3/60]
Mass flow rate = 0.02654 kg/s
mh1 + mw = mh2 + Q
0.02654(300.19 + 240) = 0.02654(441.61) + Q
Q = 2.617 Kw
Heat transfer = 2.617 Kw
Find the perpendicular distance from the point P(9,11,−8) ft to a plane defined by three points A(1,9,−4) ft, B(−4,−8,6) ft, and C(−1,−2,2) ft
Distance = ______ ft
Answer:
0 ft
Explanation:
The equation of the plane can be found from the cross product AC×BC. That vector is ...
N = (2, 11, -6) × (-3, -6, 4) = (8, 10, 21)
Then the equation of the plane is ...
8x +10y +21z = 14 . . . . . 14 = N·A
Point P satisfies this equation, so is on the plane. The distance is 0 feet.
8(9) +10(11) -8(21) = 72 +110 -168 = 14
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm^-2 . Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end.
Required:
a. How far (in miles) would this chain extend?
b. Now suppose that the density is increased to 1010 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?
Answer:
[tex]62.14\ \text{miles}[/tex]
[tex]6213727.37\ \text{miles}[/tex]
Explanation:
The distance of the chain would be the product of the dislocation density and the volume of the metal.
Dislocation density = [tex]10^5\ \text{mm}^{-2}[/tex]
Volume of the metal = [tex]1000\ \text{mm}^3[/tex]
[tex]10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}[/tex]
[tex]1\ \text{mile}=1609.34\ \text{m}[/tex]
[tex]\dfrac{10^5}{1609.34}=62.14\ \text{miles}[/tex]
The chain would extend [tex]62.14\ \text{miles}[/tex]
Dislocation density = [tex]10^{10}\ \text{mm}^{-2}[/tex]
Volume of the metal = [tex]1000\ \text{mm}^3[/tex]
[tex]10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}[/tex]
[tex]\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}[/tex]
The chain would extend [tex]6213727.37\ \text{miles}[/tex]
the pressure rise, across a pump can be expressed as where D is the impeller diameter, p, is the fluid density, w is the rotational speed, adn q is the flowrate. determine a suitable set of dimensionless parameters
Answer:
hello your question is incomplete below is the complete question
The pressure rise Δp across a pump can be expressed as Δp = f(D, p, w, Q) where D is the impeller diameter, p is the fluid density, w is the rotational speed, and Q is the flowrate. determine a suitable set of dimensionless parameters
answer : Δp / D^2pw^2 = Ф (Q / D^3w )
Explanation:
k ( number of variables ) = 5
r ( number of reference dimensions ) = 3
applying the pi theorem
hence the number of pi terms = k - r = 5 - 3 = 2
Match the use of the magnetic field to its respective description.
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Consider the flow of mercury (a liquid metal) in a tube. How will the hydrodynamic and thermal entry lengths compare if the flow is laminar
Answer:
Explanation:
Considering the flow of mercury in a tube:
When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.
Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph
Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 2.6 kg/s. If the mixture leaves the mixing chamber at 60°C.
Required:
Determine the mass flow rate of the superheated steam required.
Answer:
0.154kg/s
Explanation:
From this question we have the following information:
P1 = 300kpa
T1 = 20⁰c
M1 = 2.6kg/s
For superheated system
P2 = 300kpa
T2 = 300⁰c
M2 = ??
T2 = 60⁰c
From saturated water table
h1 = 83.91kj/kg
h3 = 251.18kj/kg
From superheated water,
h2 = 3069.6kj/kg
The equation of energy balance
m1h1 + m2h2 = m3h3
When we input all the corresponding values:
We get
m2 = -434.902/-2818.42
m2 = 0.15430
m2 = 0.154kg/s
This is the mass flow rate of the superheated steam
Please check attachment for more detailed explanation.
thank you!
This question involves the concepts of energy balance and mass flow rate.
The mass flow rate of the superheated steam required is "0.15 kg/s".
Applying the energy balance in this situation, we get:
[tex]m_1h_1+m_2h_2=m_3h_3[/tex]
where,
m₁ = mass flow rate of liquid water at 300 KPa and 200°C = 2.6 kg/s
m₂ = mass flow rate of superheated at 300 KPa and 300°C = ?
h₁ = enthalpy of liquid water at 300 KPa and 200°C = 83.91 KJ/kg (from saturated steam table)
h₂ = enthalpy of superheated at 300 KPa and 300°C = 3069.6 KJ/kg (from superheated steam table)
h₃ = enthalpy of exiting fluid at 60°C = 251.18 KJ/kg (from saturated steam table)
m₃ = mass flow rate of exiting fluid = 2.6 kg/s + m₂
Therefore,
[tex](2.6\ kg/s)(83.91\ KJ/kg)+(m_2)(3069.6\ KJ/kg)=(2.6\ kg/s+m_2)(251.18\ KJ/kg)\\m_2(3069.6\ KJ/kg-251.18\ KJ/kg)=(2.6\ kg/s)(251.18\ KJ/kg-83.91\ KJ/kg)\\\\m_2=\frac{434.902\ KW}{2818.42\ KJ/kg}[/tex]
m₂ = 0.15 kg/s
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What type of circles have two or more circles with different center points?
Answer:
Concentric circles
Explanation:
Concentric circles are two or more circles which have the same center point. The region between two concentric circles is called an annulus.
Write a program that asks the user to enter a list of numbers. The program should take the list of numbers and add only those numbers between 0 and 100 to a new list. It should then print the contents of the new list. Running the program should look something like this:
Please enter a list of numbers: 10.5 -8 105 76 83.2 206
The numbers between 0 and 100 are: 10.5 76.0 83.2
In python 3.8
nums = input("Please enter a list of numbers: ").split()
new_nums = [x for x in nums if 0 < float(x) < 100]
print("The numbers between 0 and 100 are: " + " ".join(new_nums))
When you said numbers between 0 and 100, I didn't know if that was inclusive or exclusive so I made it exclusive. I hope this helps!
. In the U.S. fuel efficiency of cars is specified in miles per gallon (mpg). In Europe it is often expressed in liters per 100 km. Write a MATLAB userdefined function that converts fuel efficiency from mpg to liters per 100 km. For the function name and arguments, use Lkm
Answer:
MATLAB Code is written below with comments in bold, starting with % sign.
MATLAB Code:
function L = Lkm(mpg)
L = mpg*1.60934/3.78541; %Conversion from miles per gallon to km per liter
L = L^(-1); %Conversion to liter per km
L = L*100; %Conversion to liter per 100 km
end
Explanation:
A function named Lkm is defined with an output variable "L" and input argument "mpg". So, in argument section, we give function the value in miles per gallon, which is stored in mpg. Then it converts it into km per liter by following formula:
L = (mpg)(1.60934 km/1 mi)(1 gallon/3.78541 liter)
Then this value is inverted to convert it into liter per km, in the next line. Then to find out liter per 100 km, the value is multiplied by 100 and stored in variable "L"
Test Run:
>> Lkm(100)
ans =
2.3522
Read through the paragraph on personal protective equipment on page 2 and fill in the blanks using the list of words below. Each word is used only once.
Sewers
Ear defenders
Replaces
Drop
Hazards
Dust mask
Hands
Employer
Safety boots
85db
Protect
Falling objects
Tread
Helmet
Gauntlets
Responsibility
Noise
PPE
I have just started a new job, and it is the responsibility of my ____________________ to supply the correct PPE for me to wear. PPE is designed to ___________________ me against workplace _____________________ . When we go on site I need good strong ______________ _____________________ to protect my feet from things like dampness, in case I ____________ something on my foot and also in case I _____________________ on an up-turned nail. My boss always asks me to use safety gloves to protect my ______________________. In fact, once I had to wear arm-length ____________________ when we were working on some _______________ |On site I also have to wear a safety ____________________ all the time. This is a good thing because it protects my head from any ___________________ ________________________. Working with other trades, such the carpenters, can be hazardous because they make lots of ______________________ with their power tools. If the noise is above ____________________ we have to use ______________ ______________________, otherwise my hearing could be damaged. Because of all the sawdust they create when they are cutting wood, they regularly have to wear a _____________ ___________________, which stops them breathing in the dust.
I have to look after all the ______________________ my boss gives me, as it is my ______________________. If something gets broken, he _______________________ it for me as he does not want to get in to trouble with the HSE.
Answer:
BESUE THE 35
Explanation:
What test should be performed on abrasive wheels
Answer:
before wheel is put on it should be looked at for damage and a sound or ring test should be done to check for cracks, to test the wheel it should be tapped with a non metallic instrument (I looked it up)
The test that should be performed on abrasive wheels is the ring test.
What is the purpose of the ring test on the abrasive wheels?The ring test can be regarded as one of the mechanical test that is used to know whether the wheel is cracked or damaged.
To carry out this test , the wheel will be arranged to be in the 45 degrees each side and it is then aligned to be at a specific diameter, this can be done by the expert in this field to know the state of that wheel.
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You are playing guitar on a stool that is 22" tall. How tall is the stool if it is expressed as a combination of feet and inches?
Answer:
1 foot 10 inches
Explanation:
1 foot = 12 inches + 10 inches = 22 inches
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Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more electrical resistance it has. Who is correct?
Answer:
Both of them are wrong
Explanation:
The two technicians have given the wrong information about the wires.
This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.
Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance
So this practically makes the second technician wrong too
Describe how the fracture behavior would be different for a fiber-reinforced tape such as duct tape.
Answer:
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Explanation:
The fracture behavior would be different for a fiber-reinforced tape in the following way :
* It's behavior during tensile stress and its fracture behavior.
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______
Answer:
Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]
Explanation:
Radius of Gaussian surface = R
Charge in the Sphere ( Gaussian surface ) = Q
lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R
To determine the net electric flux through the Gaussian surface
we have to apply Gauci law
Ф = 4[tex]\pi r^2 E[/tex]
Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]
= [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]
A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
of the water when the specific heat capacity of water is 4200 J/kg °C.
Answer:
2.728 kg
Explanation:
The units help you keep the calculation straight.
[tex]\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}[/tex]