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
Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
Hope this helps...........
Match the use of the magnetic field to its respective description.
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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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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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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
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.
Learn more about ring test on:
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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
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 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]
. 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
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]
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
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