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]
Design a circuit that outputs a 1 when the bit pattern (101) has been applied to input, and 0 otherwise.
Answer:
see the attachment
Explanation:
The circuit shown uses Nand and Nor gates to produce the desired logic. The input bits are numbered 0 to 2, right to left.
The logic is ...
out = ((b2·b0)' + b1)'
out = b2·b1'·b0
when you reach a yield sign yield to cross traffic and ___ before you enter the intersection
Answer:
Wain for a safe gap
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
Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools. Technician B states that you should always use impact sockets with impact guns. Who is correct?
Answer:
Both Technician A and Technician B are correct
Explanation:
Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.
Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.
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.
When jump-starting a car, always remember ___________
Answer:
to keep the cables away from any belts so that when starting the car back up, they do not get tangled or caught up into the engine
Explanation:
When jump-starting a car, always remember jumper cables typically have two clamps, one with the label “positive” in red and “negative” in black.
What is a jump-start?Jump-starting a car means starting a car when it is off for a long time, and it should be started by giving heat to a battery. The cables of the car are attached to the battery box and the current is given.
The steps to jump-starting a car are:
With a functional battery, start the vehicle. Start a working car's engine, then move it near to the vehicle with the dead battery.Attach jumper cords to the batteries.Start the vehicle.Disconnect the two batteries with care.After the jump start, check the cables' condition.Thus, when jump-starting a car, always remember the cables are in the right position.
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Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of 5.12 MPa, compute the critical resolved shear stress.
Answer:
imma leabe
Explanation:
leabe leave*
What are some of the trade-offs of a move to an enterprise-level analytics solution for individual end users who might have grown accustomed to working with their own customized solutions for generating data?
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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Equal moles of pure liquid 1-Butanol, Benzene, and Phenol form an ideal solution system at 353K. Determine Yi for each component at vapor-liquid equilibrium (VLE) ofthe mixtureat 353K.
Answer:
[tex]\mathbf{y_1 =0.1750}[/tex]
[tex]\mathbf{y_2 = 0.8088}[/tex]
[tex]\mathbf{y_3 = 0.0161}[/tex]
Explanation:
Given that:
The temperature of the ideal solution formed by the compounds = 353K
= (353 - 273)° C
= 80° C
The data below shows the Antoine constant obtained for 1-Butanol, Benzene, and Phenol.
Compound A B C
1 - Butanol 15.3144 3212.43 182.739
Benzene 13.7819 2726.81 217.572
Phenol 14.4397 3507.80 175.400
By the application of the Antoine constant, we can find the vapor pressure of each corresponding component at the given temperature of 80° C.
The Antoine equation is expressed as:
[tex]In (P^*) = A - \dfrac{B}{T+C}[/tex]
[tex]P_1 ^* = exp \bigg [15.314 - \dfrac{3212.43}{80+182.739} \bigg ][/tex]
[tex]P_1 ^* = exp \bigg [15.314 - \dfrac{3212.43}{262.739} \bigg ][/tex]
[tex]P_1 ^* = 21.9267 \ kPa[/tex]
[tex]P_2^* = exp \bigg [13.7819 - \dfrac{2726.81}{80+217.572} \bigg ][/tex]
[tex]P_2^* = exp \bigg [13.7819 - \dfrac{2726.81}{297.572} \bigg ][/tex]
[tex]P_2^* = 101.3287 \ kPa[/tex]
[tex]P_3^* = exp \bigg [14.4387 - \dfrac{3507.80}{80+175.400} \bigg ][/tex]
[tex]P_3^* = exp \bigg [14.4387 - \dfrac{3507.80}{255.4} \bigg ][/tex]
[tex]P_3^* =2.0221 \ kPa[/tex]
However, since the ideal solution has equimolar composition.
Then:
component (1) = 1 mole ,
component (2) = 1 mole,
component (3) = 1 mole,
The total mole = 1 + 1 + 1 = 3,
Thus, mole fraction = mole of component/total mole,
mole fraction of component (1) [tex]x_1[/tex] = 1/3,
mole fraction of component (2) [tex]x_2[/tex] = 1/3,
mole fraction of component (3) [tex]x_3[/tex] = 1/3,
i.e.
[tex]x_1 =x_2=x_3 = \dfrac{1}{3}[/tex]
Using Raoult's law, The total pressure is computed as:
[tex]P = x_1P_1^* + x_2 P_2^* + x_3P_3^*[/tex]
[tex]P = \dfrac{1}{3}(21.9267) + \dfrac{1}{3} (101.3287) + \dfrac{1}{3} (2.0221)[/tex]
P = 41.7592 kPa
and;
[tex]P_1 *x_1 = y_1P[/tex]
[tex]\implies y_1 = \dfrac{P_1\times x_1}{P}[/tex]
Thus:
[tex]y_1 = \dfrac{21.9267 \times \dfrac{1}{3}}{41.7592}[/tex]
[tex]\mathbf{y_1 =0.1750}[/tex]
[tex]y_2 = \dfrac{P_2\times x_2}{P}[/tex]
[tex]y_2 = \dfrac{101.3287 \times \dfrac{1}{3}}{41.7592}[/tex]
[tex]\mathbf{y_2 = 0.8088}[/tex]
[tex]y_3 = \dfrac{P_3\times x_3}{P}[/tex]
[tex]y_3 = \dfrac{2.0221\times \dfrac{1}{3}}{41.7592}[/tex]
[tex]\mathbf{y_3 = 0.0161}[/tex]
why do u have to have certain limits for questions
Answer:
ok
Explanation:
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
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
which of the following is a function of a safety device
Answer:
what are the options available?
Sharon has just invented a new tractor that will plow and plant a new hybrid of corn at the same time. Which type of engineer is she?
Answer:
Agricultural engineer
Available for ceiling and wall installation, can be covered with a finish covering material, plastered, or mounted directly to the ceiling or wall. TERE SESELT BP
A. hydronic heating systems
B. evaporative systems
C. electric radiant heating
D. panels unit cooling systems
Answer:
Hydronic heating system
Explanation:
From the available options, Hydronic heating systems can be covered with a finishing covering material, plastered or can be mounted directly to the celing or wall.
This system heats water and then moves it through pipes that are sealed pipes to radiators all around a house.
This sealed system is useful for heating towel rails, floor slabs, and also swimming pools, whenever it is useful
Hydronic Heating water is heated through the use of super energy
1) What conditions must be satisfied to assure that a refutable hypothesis will be obtained just by examining the structure of the FOC? (Assume that SOCs are satisfied as well).
Answer:
The derivative of a function must be equal to zero ( 0 ) if this condition is not met then a hypothesis is refuted
Explanation:
The conditions that must be satisfied to assure that a refutable hypothesis will be obtained by examining the structure of the FOC is
The derivative of a function must be equal to zero ( 0 ) if this condition is not met then a hypothesis is refuted
i.e. dy/dx w.r.t.x = 0 for FOC to be refutable
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air at the inlet is 1.20 kg/m^3 and 0.95kg/m^3 at the exit, determine the percent increase in the velocity of the air as it flows through the dryer.
Answer:
% increase = 26.32%
Explanation:
From conservation of mass, we can say that;
Mass flow rate at inlet = mass flow rate at exit.
Thus;
m'1 = m'2
Formula for mass flow rate is;
m' = ρV'
Where V' is volumetric flow rate = Av
Thus;
m' = ρAv
Where;
ρ is density
A is area
v is velocity
Therefore from m'1 = m'2, we can say that;
ρ1•A1•v1 = ρ2•A2•v2
Since the duct has a constant diameter, then A1 = A2
Thus, we now have;
ρ1•v1 = ρ2•v2
Making v2 the subject, we have;
v2 = ρ1•v1/ρ2
Now, since we want to find the percent increase in the velocity of the air as it flows through the dryer,we would use;
% increase = ((v2 - v1)/v1) × 100%
We have v2 = ρ1•v1/ρ2
Thus;
% increase = ((ρ1•v1/ρ2) - v1)/v1) × 100%
Factorizing v1 out, we have;
% increase = ((ρ1/ρ2) - 1)/1) × 100%
We are given;
ρ1 = 1.2 kg/m³
ρ2 = 0.95 kg/m³
Thus;
% increase = ((1.2/0.95) - 1)/1) × 100%
% increase = 26.32%
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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A specimen of aluminum having a rectangular cross section 10 mm x 12.7 mm is pulled in tension with 35,500 N (8000 lbf) force, producing only elastic deformation. Calculate the resulting strain.
Answer:
The resulting strain is 4.05 x 10⁻³
Explanation:
Given;
dimension of the specimen = 10 mm x 12.7 mm
Cross sectional area of the aluminum specimen = 10 mm x 12.7 mm = 127 mm² = 1.27 x 10⁻⁴ m²
applied force, F = 35,500 N
Young's modulus is given by;
[tex]E = \frac{Stress}{Strain}\\\\E = \frac{F}{A(strain)}\\\\E = \frac{F}{A(\epsilon)}\\\\\epsilon = \frac{F}{A E}\\\\[/tex]
Where;
ε is the resulting strain
E is Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
[tex]\epsilon = \frac{35500}{(1.27*10^{-4}) (69*10^{9})}\\\\ \epsilon = 4.05*10^{-3}[/tex]
Therefore, the resulting strain is 4.05 x 10⁻³
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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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
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a feed of 0.02 ipr? What is the cutting time?
Answer:
a) the metal removal rate is 14.4 in³/min
b) the cutting time is 0.98 min
Explanation:
Given the data from the question
first we find the rpm for the spindle of the drilling tool, using the equation
Ns = 12V/πD
V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)
so we substitute
Ns = 12 × 120 / π2
Ns = 1440 / 6.2831
Ns = 229.18 rmp
Now we find the metal removal rate using the equation
MRR = (πD²/4) Fr × Ns
Fr is the feed rate( 0.02 ipr ),
so we substitute
MRR = ((π × 2²)/4) × 0.02 × 229.18
MRR = 14.3998 ≈ 14.4 in³/min
Therefore the metal removal rate is 14.4 in³/min
Next we find the allowance for approach of the tip of the drill
A = D/2
A = 2/2
= 1 in
now find the time required to drill the hole
Tm = (L + A) / (Fr × Ns)
Lis the the depth of the hole( 3.5 in)
so we substitute our values
Tm = (3.5 + 1) / (0.02 × 229.18 )
Tm = 4.5 / 4.5836
Tm = 0.98 min
Therefore the cutting time is 0.98 min
Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
Hope this helps...........
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]
Match the use of the magnetic field to its respective description.
oooExplanation:
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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
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]
List in order first three steps to square a board
Answer:
STEP1 Cut to Rough Length
STEP2 Cut to Rough Width
STEP 3 Face-Jointing
HOPE THAT HELPSSS!!!
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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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
