According to the building-up principle or Aufbau principle, the subshell is typically filled next after the 3d subshell is the 4p subshell. The correct option is A).
What are subshells?A subshell is the collection of states within a shell that are indicated by the azimuthal quantum number, l. S, P, D, and F subshells are represented by the values l = 0, 1, 2, and 3, respectively.
The formula for the maximum number of electrons that can fit into a subshell is 2(2l + 1). The highest level of probability for an electron to occur is in an atomic orbital.
The route that electrons travel while navigating the restrictions of the shell is referred to as a subshell. There are four different subshell classifications. The letters s, p, d, and f are used to identify the subshells.
Therefore, the correct option is A) 4p.
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Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperature of 80oC. Calculate the average heat transfer coefficient and the heat transfer rate per meter length of pipe.\
Answer:
Average heat transfer coefficient = 31 kw/m^2 k
Heat transfer rate per meter length of pipe = 116.808 KW
Explanation:
water temperature = 20⁰c,
free-stream velocity = 1.5 m/s
circular pipe diameter = 2.0 cm = 0.02 m
surface temperature = 80⁰c
A) calculate average heat transfer coefficient
we apply the formula below :
m = αAv
A (area) = [tex]\pi /4 (d)^2[/tex]
m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5
= 10^3 * 0.7857( 0.0004) * 1.5
= 0.4714 kg/s
Average heat transfer coefficient
h = [tex]\frac{m(cp)}{A}[/tex] , A = [tex]\pi DL[/tex]
L = 1 m , m = 0.4714 kgs , cp = 4.18
back to equation
h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex] = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k
B) Heat transfer rate per meter length of pipe
Q = ha( ΔT ), a = [tex]\pi DL[/tex]
= 31 * 0.0628 * ( 80 - 20 )
= 31 * 0.0628 * 60 = 116.808 KW
An 8-m long, uninsulated square duct of cross section 0.2m x 0.2m and relative roughness 10^-3 passes through the attic space of a house.. Hot air (80°C) enters an 8 m long un-insulated square duct (cross section 0.2 m x 0.2 m) that passes through the attic of a house at a rate of 0.15 m^3 /s. The duct is isothermal at a temperature of 60°C. Determine the rate of heat loss from the duct to the attic space and the pressure difference between the inlet and outlet sections of the duct.
Answer:
the rate of heat loss from the duct to the attic space = 1315.44 W
the pressure difference between the inlet and outlet sections of the duct = 7.0045 N/m²
Explanation:
We know that properties of air 80⁰C and 1atm (from appendix table) are;
density p = 0.9994 kg/m³, Specifice heat Cp = 1008 J/kg.⁰C
Thermal conductivity k = 0.02953 W/m.⁰C, Prandtl number Pr = 0.7154,
Kinematic viscosity v = 2.097 × 10⁻⁵ m²/s
haven gotten that, we calculate the hydraulic diameter of square duct
Dh = 4Ac / P { Ac = is cross sectional area of duct and P = perimeter}
now we substitute a² for Ac and 4a for P ( we know from the question that a = 0.2 m)
Dh = 4a² / 4a
Dh = 4(0.2)² / 4(0.2)
Dh = 0.2 m
Now we calculate the average velocity of air
Vₐ = Vˣ / Ac { vˣ = volume flow rate of air}
Vₐ = Vˣ / a² { Ac = a² }, we know that a = 0.2m₂, Vˣ = 0.15 m³
Vₐ = 0.15 / (0.2)²
Vₐ = 3.75 m/s
Next we calculate the Reynolds number
Re = Vₐ Dh / V
Re = (3.75 × 0.2) / 2.097× 10⁻⁵
Re = 35765.379
The Reynolds number IS GREATER than 10,000
so the flow is turbulent and entry length in this case is nearly 10 times the hydraulic diameter
Lh ≈ Lt ≈ 10D
= 10 × 0.2
= 2m
As this length is quite small when compared to the total of tube, we assume fully developed flow for the entire tube length.
Now we calculate the Nusselt number from this relation;
Nu = 0.023 Re⁰'⁸ Pr⁰'³
so we substitute for Re and Pr
Nu = 0.023(35765.379)⁰'⁸ (0.7154)⁰'³
Nu = 91.4
Now calculate the convective heat transfer coefficient
h = Nu × K/ Dh
we substitute
h = 91.4 × 0.02953 W/m.°C / 0.2 m
h = 13.5 W/m².°C
We calculate the surface area of the square duct
Aₓ = 4aL { L= length of duct}
we substitute
Aₓ = 4 × 0.2 × 8
Aₓ = 6.4 m²
Mass flow rate of air
m = pVˣ
we substitute again ( from our initials)
m = 0.9994 kg/m₃ × 0.15 m³/s
m= 0.150 kg/s
We calculate the exit temperature of the air from the duct
Te = Ts - (Ts -Ti) exp ( - hAₓ / mCp)
we know that
Ts = 60°C , Ti = 80°C, h = 13.5 W/m².°C , Aₓ = 6.4m², m = 0.150 kg/s , Cp = 1008 J/kg.°C
we substitute
Te = 60 - (60-80) exp(- ((13.5 × 6.4)/(0.15 × 1008))
Te = 71.3°
Now we calculate the rate of heat loss from the duct.
Q = mCp ( Ti -Te )
we substitute again
Q = 0.150 × 1008 × ( 80 - 71.3 )
Q = 1315.44 W
Next we calculate the estimated friction factors by using Haaland equation
1/√f = - 1.8log₁₀ [ 6.9/Re + (E/D)/3.7)¹'¹¹]
we know that E/D = relative roughness = 10⁻³
we substitute
so
1/√f = - 1.8log₁₀ [ (6.9/35765.379) + ( 10⁻³/3.7)¹'¹¹]
1/√f = - 1.8log₁₀ { 0.000192924 + 0.00010947}
1/√f = - 1.8log₁₀ 0.000302324
√f = 1/6.334
f = (1/6.334)²
f = 0.02492
We calculate the pressure difference between inlet and outlet sections of the duct
ΔPl = fLPVa² / Dh × 2
ΔPl = {0.02492 × 8 × 0.9994 × (3.75)²} / 0.2 × 2
ΔPl = 2.8018 / 0.4
ΔPl = 7.0045 N/m²
Therefore pressure deference is 7.0045 N/m²
A cruise missile under test is moving horizontally at Ma =2 in the atmosphere at an elevation of 2000 m (Air temperature is 2 °C). Determine the half-angle "alpha" of the Mach cone.
Answer: the half-angle "alpha" of the Mach cone = 30⁰
Explanation:
To calculate the half-angle "alpha" of the Mach cone.
we say ;
Sin∝ = 1 / Ma
given that Ma = 2
now we substitute
Sin∝ = 1 / 2
Sin∝ = 0.5
∝ = Sin⁻¹ 0.5
∝ = 30⁰
Therefore, the half-angle "alpha" of the Mach cone is 30⁰
A single-threaded power screw is 35 mm in diameter with a pitch of 5 mm. A vertical load on the screw reaches a maximum of 5 kN. The coefficients of friction are 006 for the collar and 009 for the threads, while the frictional diameter of the collar is 45 mm. Find the overall efficiency and the torque to raise and lower the load for
Answer:
the torque required to RAISE the load is Tr = 18.09 Nm
the torque required to LOWER the load is Tl = 10.069 ≈ 10.07 Nm
the Overall Efficiency e = 0.2199 ≈ 0.22
Explanation:
Given that; F = 5 kN, p = 5mm, d = 35mm
Dm = d - p/2
Dm = 35 - ( 5/2) = 35 - 2.5
DM = 32.5mm
So the torque required to RAISE the load is
Tr = ( 5 × 32.5)/2 [(5 + (π × 0.09 × 32.5)) / ( (π × 32.5) - ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]
Tr = 81.25 × (14.1892 / 101.6518) + 6.75
Tr = 11.3414 + 6.75
Tr = 18.09 Nm
the torque required to LOWER the load is
Tl = ( 5 × 32.5)/2 [(π × 0.09 × 32.5) - 5) / ( (π × 32.5) + ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]
Tl = 81.25 × 4.1892 / 102.5518 + 6.75
Tl = 3.3190 + 6.75
Tl = 10.069 ≈ 10.07 Nm
So since torque required to LOWER the load is positive
that is, the thread is self locking
Therefore the efficiency is
e = ( 5 × 5 ) / ( 2π × 18.09 )
e = 25 / 113.6628
e = 0.2199 ≈ 0.22
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.
Answer:
The answer is below
Explanation:
1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:
[tex]n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm[/tex]
2) The speed of the rotor is the motor speed. The slip is given by:
[tex]Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm[/tex]
3) The frequency of the rotor is given as:
[tex]f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz[/tex]
4) At standstill, the speed of the motor is 0, therefore the slip is 1.
The frequency of the rotor is given as:
[tex]f_r=slip*f_s\\f_r=1*60=60\ Hz[/tex]
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as
Answer:
Explanation:
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:
[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )
where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.
Q1) Determine the force in each member of the
truss and state if the members are in tension or
compression.
Set P1 = 10 kN, P2=15 KN
Answer:
CD = DE = DF = 0BC = CE = 15 N tensionFA = 15 N compressionCF = 15√2 N compressionBF = 25 N tensionBG = 55/2 N tensionAB = (25√5)/2 N compressionExplanation:
The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.
At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.
Then link CE has 15 N tension.
The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.
In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.
At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.
At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.
The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.
In summary, the link forces are ...
(25√5)/2 N compression in AB15 N tension in BC25 N tension in BF0 N in CD, DE, and EF15 N tension in CE15√2 compression in CF15 N compression in FA_____
Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.
Air at 1 atm, 15°C, and 60 percent relative humidity is first heated to 20 °C in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 25°C and 65 percent relative humidity. Determine:
a. the amount of steam added to the air.
b. the amount of heal transfer to the air in the heating section.
Answer: a = change in w =0.00656
b = q = 5.1kj/kg
Explanation:
Find explanation in the attached file
The amount of steam added to the air a = change in w =0.00656 b = q = 5.1kj/kg
What is steam?The digital game retail and distribution service Steam is provided by Valve. In order to allow Valve to automatically update its games, it was first released as a software client in September 2003. In late 2005, it was expanded to include the distribution and sale of games from other publishers.
a) We can use the absolute humidity we and wg to determine the amount
of moisture added Aw.
Aww3-W2
Aw= 0.01291 -0.00635
Aw= 0.00656
b) To determine the heat transfer q we will need the enthalpies h and h2.
kJ
kg 9 = 36.2
kJ kg
31.1
q=5.1
kJ
kg
RESULT
Do = 0.00656
kJ
kg
9 = 5.1
Therefore, The amount of steam added to the air a = change in w =0.00656 b = q = 5.1kj/kg
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how to build a laser pointer?
Answer:
It's easier to buy one, but you can search for a tutorial on how to make one on Yuotube.
There are quite a few videos on how to make one.
Based on the hardness values determined in Part 1, what is the tensile strength (in MPa) for each of the alloys?
(a) 0.25 wt%C with spheroidite,
(b) 0.25 wt%C with coarse pearlite,
(c) 0.60 wt%C with fine pearlite, and
(d) 0.60 wt%C with coarse pearlite.
The tolerance to all answers is +/-10 %.
Answer:
a. 115
b. 135
c. 220
d. 185
Explanation:
Spheriodite is microscopic constituents in some steels which is composed of spherically shaped cementide particle. It is most ductile and softest type of steel. Pearlite is two phased lamellar compose of alternating layer of ferrite and cementite. It is hard and strong but not tough. It is applied on cutting tools like chopper, blades and knives.
Tech A says that a transistor has a single P–N junction. Tech B says that a transistor is a semiconductor device used as a switch and to amplify currents. Who is correct? Group of answer choices
Answer:
Both Technician A and technician B are correct.
Explanation: A transistor must have a P-N junction as that is where the positive and negative charges are connected.
A transistor also can be described as a semiconductor which acts as a switch and can be used to amplify currents. Transistors are very key and vital to electronic devices especially the mobile phones in recent times, it helps to ensure that electronic systems perform optimally.
The charges in the P-N junction is controlled by the availability of Positive and negative electrons.
Write about traditional brick production in Pakistan
Answer:
Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.
Explanation:
In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about 1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.
There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.
Assume that the heat is transferred from the cold reservoir to the hot reservoir contrary to the Clausis statement of the second law. Prove that this violates the increase of entropy principle—as it should according to Clausius.
Answer: hello attached below is the diagram which is part of your question
Total entropy change = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k it violates Clausius increase of entropy which is Sgen > 0
Explanation:
Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .
applying the increase in entropy principle to prove this
temp of cold reservoir (t hot)= 600 k
temp of hot reservoir(t cold) = 1220 k
energy (q) = 100 kj
total entropy change = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k
entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k
entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k
hence it violates Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0
A plate is supported by a ball-and-socket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem?
Answer:
There are five (5) unknown support reactions in this problem.
Explanation:
A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.
A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.
A ball-and-socket joint have reaction forces in all 3 cardinal directions. This allows it to move in the x-y-z plane.
The total unknown reactions on the member are five in number.
Suppose that the resistors in the various circuit diagrams represented the resistances of lightbulbs. When a lightbulb "burns out," the circuit is open through that particular component, that is R is infinite. Would the remaining bulbs continue to burn?
Answer:
The Remaining Bulbs will either burn out( draw more current ) or Not burn out depending on the arrangement of the bulbs in the circuit
Explanation:
The Remaining Bulbs will either burn more brightly ( draw more current ) or Burn less brightly ( draw less current).and this depends on the arrangement of the light bulbs in the various circuit.
If the light bulbs are connected in series the remaining bulbs will burn out as soon as any light bulb burns out and this is because bulbs connected in series receive the same amount of current ,
If the light bulbs are connected in parallel the remaining bulbs will not burn out because each bulb receives current based on its resistance.
Assume you have created a class named MyClass and that is contains a private field named
myField and a nonstatic public method named myMethod(). Which of the following is
true?
a. myMethod() has access to and can use myField
b. myMethod() does not have access to and cannot use myFeild.
c. myMethod() can use myField but cannot pass it to other methods.
d. myMethod() can use myField only in myField is passed to myMethod() as a
parameter.
Answer:
a. myMethod() has access to and can use myField.
Explanation:
Logic programming is a kind of programming which is largely based on formal logic. The statement are written in logical forms which express rules about the domain. In the given scenario the my method will have access to my field which is private field. My method non static public field can also use my field class.
Technician A says that when the malfunction indicator light or service engine light is on you should retrieve the diagnostic trouble code and follow the manufacturers recommended procedure. Technician B says that all obd-ll monitors must have the enabled criteria achieved before a test is performed. Who is correct?
Answer:both
Explanation:
what scale model proves the initial concept?
Answer: A prototype
Explanation:
The scale model that proves the initial concept is called a domain model.
What is a scale model?A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.
A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.
A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.
Therefore, a domain model is the scale model that proves the initial concept.
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A permanent-magnet dc motor has the following parameters: Ra = 0.3 Ω and kE = kT = 0.5 in MKS units. For a torque of up to 10 Nm, plot its steady state torque-speed characteristics for the following values of Va: 100 V, 75 V, and 50 V.
Answer:
load speeds:
For V = 100 v speed = 188 rad/sec
For V = 75 v speed = 138 rad/sec
For V = 50 v speed = 88 rad/sec
Explanation:
Given data
Ra = 0.3 Ω
Ke = Kt = 0.5
torque = 10 Nm
using a constant torque = 10 Nm we can calculate the various load speed for the given values of 100 v , 75 v, 50 v
attached below is the detailed solutions and plot
A step-up transformer has an input voltage of 110 V (rms). There are 100 turns on the primary and 1500 turns on the secondary. What is the output voltage?
Answer:
V2= 1666.6 volts
Explanation:
Given data
primary turns N1= 100 turns
secondary turns N2= 1500 turns
primary voltage V1= 110 volts
secondary voltage V2= ?
We can solve for the output voltage using the turns ration sated below
Turns Ratio = N1 / N2 = V1 / V2
Substituting our given data into the expression we have
100/1500= 110/V2
Making V2 subject of formula we have
V2= 110/(100/1500)= 1666.6 volts
V2= 1666.6 volts
Hence the secondary voltage is 1666.6 volts
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
Answer:
As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.
In a series motor, the field electromagnet consists of A) a winding connected in parallel with the armature. B) a winding connected in parallel with the armature and a second winding connected in series with the armature. C) a winding connected in series with the armature. D) a winding connected in series with a separate dc power source.
Answer:
C) a winding connected in series with the armature.
Explanation:
In a series motor, an electromagnet is used as a stator to generate its magnetic field. The field coil of this stator are connected through a commutator in series with the rotor windings. This stator which is the armature windings will conduct AC even on a DC machine, due to the periodically reverses current direction (commutation) or due to electronic commutation (as in brushless DC motors).
Define centrifugal pump. Give the construction and working of centrifugal pump.
A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observed to vibrate with a natural frequency of 9.3 HZ. What is the moment of inertia of the beam's cross section about its neutral axis?
Answer:
moment of inertia = 4.662 * 10^6 [tex]mm^4[/tex]
Explanation:
Given data :
Mass of machine = 400 kg = 400 * 9.81 = 3924 N
length of span = 3.2 m
E = 200 * 10^9 N/m^2
frequency = 9.3 Hz
Wm ( angular frequency ) = 2 [tex]\pi f[/tex] = 58.434 rad/secs
also Wm = [tex]\sqrt{\frac{g}{t} }[/tex] ------- EQUATION 1
g = 9.81
deflection of simply supported beam
t = [tex]\frac{wl^3}{48EI}[/tex]
insert the value of t into equation 1
W[tex]m^2[/tex] = [tex]\frac{g*48*E*I}{WL^3}[/tex] make I the subject of the equation
I ( Moment of inertia about the neutral axis ) = [tex]\frac{WL^3* Wn^2}{48*g*E}[/tex]
I = [tex]\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}[/tex] = 4.662 * 10^6 [tex]mm^4[/tex]
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget that is 5.5 mm in diameter and 5.0 mm thick after 0.3 sec welding time. The unit melting energy for a certain sheet metal is 9.5 J/mm3 . The electrical resistance between the surfaces is 140 micro ohms, and only one third of the electrical energy generated will be used to form the weld nugget (the rest being dissipated), determine the minimum current level required.
Answer:
minimum current level required = 8975.95 amperes
Explanation:
Given data:
diameter = 5.5 mm
length = 5.0 mm
T = 0.3
unit melting energy = 9.5 j/mm^3
electrical resistance = 140 micro ohms
thickness of each of the two sheets = 3.5mm
Determine the minimum current level required
first we calculate the volume of the weld nugget
v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3
next calculate the required melting energy
= volume of weld nugget * unit melting energy
= 118.73 * 9.5 = 1127.94 joules
next find the actual required electric energy
= required melting energy / efficiency
= 1127 .94 / ( 1/3 ) = 3383.84 J
TO DETERMINE THE CURRENT LEVEL REQUIRED use the relation below
electrical energy = I^2 * R * T
3383.84 / R*T = I^2
3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2
therefore 8975.95 = I ( current )
Assume a regulator has a percent load regulation of 0.5%. What is the output voltage at full-load if the unloaded output is 12.0 V
Answer:
11.94 V
Explanation:
Generally the regulated voltage drops as load increases. When the voltage has dropped by 0.5%, it will be 60 mV less than the nominal value:
12.0 V - 0.06 V = 11.94 V . . . . full load voltage
Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 0.8m and 7m/s, respectively. Determine (a) the flow depth and the Froude number after the jump (b) the head loss (c) the dissipation ratio.
Answer:
A) Flow depth = 2.46 m, Froude number after jump = 0.464
B) head loss = 0.572 m
C) dissipation ratio = 0.173
Explanation:
Given data :
Velocity before jump ( v1 ) = 7 m/s
flow depth before jump ( y1 ) = 0.8 m
g = 9.81 m/s
Esi = 3.3 m ( calculated )
attached below is a detailed solution of the problem
Strain gage is a device that senses the strain of the structure. The property of the strain gage that is used to correlate with the strain to be measured is
Answer:
resistance
Explanation:
A strain gauge changes resistance with applied strain.
Conductivity is the reciprocal of what?
Answer:
Electrical conductivity or specific conductance is the reciprocal of electrical resistivity. It represents a material's ability to conduct electric current. It is commonly signified by the Greek letter σ (sigma), but κ (kappa) (especially in electrical engineering) and γ (gamma) are sometimes used.
The cold drawn AISI 1040 steel bar with 25-mm width and 10-mm thick has a 6- mm diameter thru hole in the center of the plate. The plate is subjected to a completely reversed axial load that fluctuates from 12kN to 28kN. Use notch sensitivity of 0.83.
Required:
a. Estimate the fatigue factor of safety based on yielding criteria.
b. Estimate the fatigue factor of safety based on Goodman and Morrow criteria.
Answer:
A) ( N ) = 1.54
B) N ( Goodman ) = 1.133, N ( Morrow) = 1.35
Explanation:
width of steel bar = 25-mm
thickness of steel bar = 10-mm
diameter = 6-mm
load on plate = between 12 kN AND 28 kN
notch sensitivity = 0.83
A ) Fatigue factor of safety based on yielding criteria
= δa + δm = [tex]\frac{Syt}{n}[/tex] = 91.03 + 227.58 = 490 / N
therefore Fatigue number of safety ( N ) = 1.54
δa (amplitude stress ) = kf ( Fa/A) = 2.162 * ( 8*10^3 / 190 ) = 91.03 MPa
A = area of steel bar = 190 mm^2 , Fa = amplitude load = 8 KN , kf = 2.162
δm (mean stress ) = kf ( Fm/A ) = (2.162 * 20*10^3 )/ 190 = 227.58 MPa
Fm = mean load = 20 *10^3
B) Fatigue factor of safety based on Goodman and Morrow criteria
δa / Se + δm / Sut = 1 / N
= 91.03 / 183.15 + 227.58 / 590 = 1 /N
Hence N = 1.133 ( based on Goodman criteria )
note : Se = endurance limit (calculated) = 183.15 , Sut = 590
applying Morrow criteria
N = 1 / ( δa/Se) + (δm/ δf )
= 1 / ( 91.03 / 183.15 ) + (227.58 / 935 )
= 1.35
