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.
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
The magnitude of the normal acceleration is
A) proportional to radius of curvature.
B) inversely proportional to radius of curvature.
C) sometimes negative.
D) zero when velocity is constant.
Answer:
b. inversely proportional to radius of curvature
Explanation:
In curvilinear motions, the normal acceleration which is also called the centripetal acceleration is always directed towards the center of the circular path of motion. This acceleration has a magnitude that is directly proportional to the square of the speed of the body undergoing the motion and inversely proportional to the radius of the curvature of the motion path. The centripetal or normal acceleration a, can be given by;
a = [tex]\frac{v^2}{r}[/tex]
Where;
v = speed of the body
r = radius of curvature.
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:
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.
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
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.
To learn more about the scale model, refer to the below link:
https://brainly.com/question/14341149
#SPJ2
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.
Air at 30 C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 5 C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0
Answer:
A) The relative humidity : 0.818 (81.8%), Temperature at C = 14.4⁰c
B) The rate of energy destruction = 0.0477 kw
Explanation:
Given data :
at point 1 : m1 = 3 kg/min , T1 = 30⁰c, p1 = 1 bar, ∅ = 0.50 ( 50%)
at point 2 : T2 = 5⁰c, P2 = 1 bar, m2 = 5 kg/min
at point 3 : p3 = 1 bar
What improves the structured approach in design?
A team is adopting a structured approach in design which helps them to improve the ___ of the design.
Answer:
efficiency
Explanation:
Answer:
The correct answer is Efficiency.
Explanation:
I got it right on the plato test.
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capacitor and a 301 Ω resistor.
(a) What is the impedance of the circuit?
(b) What is the rms current through the resistor?
(c) What is the average power dissipated in the circuit?
(d) What is the peak current through the resistor?
(e) What is the peak voltage across the inductor?
(f) What is the peak voltage across the capacitor?
The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
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.
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).
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.
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 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
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 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]
A student lab group is brainstorming the design of an experiment that uses an ammeter (measures current) and different resistors to determine the effect of the resistance of a resistor upon the current in a simple circuit. Which Post-it note describes the most effective design?Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data. Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Move the ammeter to a different location in the circuit. Measure the current at this new location. Continue moving the ammeter to different locations within the circuit but be careful to keep the resistor in a fixed location. Measure and record all current values. Obtain a variety of batteries and build several circuits. Make sure that each circuit has at least one resistor and make sure that the resistance values are different in the different circuits. Place various ammeters in each circuit. Measure the number of batteries and the current for each of the circuits. Record the resistance values used in each of these circuits. Put a 10.0-ohm resistor in a circuit with a single D-cell. Measure the current in the circuit. Add a second D-cell and measure the current with two D-cells. Repeat trials for three, four, and five D-cells, being careful to get accurate current measurements for a fixed amount of resistance in each trial.
Answer:
Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.
Explanation:
The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.
A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 sec/mm2 in Chvorinov's formula to calculate the solidification time.
Required:
a. Calculate the minimum time (minutes) for the casting to solidify.
b. Discuss if the result in part (a) is the same when casting grey cast iron.
Answer:
a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min
b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)
Explanation:
Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²
first we find the volume and Area;
Volume V = πD²t / 4
Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³
Area A = 2πD²/ 4 + πDt
Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}
Area A = 392,699.08 + 31,415.93
Area A = 424,115 mm²
a)
Chvorinov’s rule
T(aluminium) = Cm (V/A) ²
T(aluminium) = 2.0 × (3,926,991 / 424,115) ²
T(aluminium) = 171.5 s = 2.86 min
∴ the minimum time (minutes) for the casting to solidify is 2.86 min
b)
For cast iron
Cm (mold constant = 1.488 sec/mm²)
Chvorinov’s rule
T(iron) = Cm (V/A) ²
T(iron) = 1.488 × (3,926,991 / 424,115) ²
T(iron) = 127.5720s = 2.13 min
Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)
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
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 )
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 after jump = 2.46 m, Froude number after jump = 0.464
B) head loss = 0.572 m
C) dissipation ratio = 0.173
Explanation:
Given data :
width of channel = 10-m
velocity of before jump (V1) = 7 m/ s
flow depth before jump (y1) = 0.8 m
A) determining the flow depth and the Froude's number after the jump
attached below is the solution
B) head loss
HL = Y1 -Y2 + [tex]\frac{V_{1} ^2 - V_{2} ^2}{2g}[/tex] = 0.8 - 2.46 + [tex]\frac{49 - 5.1984}{19.62}[/tex] = 0.572
C) dissipation ratio
HL / Es1 = 0.572 / 3.3 = 0.173
Es1 = 0.8 + [tex]\frac{7^2}{2*9.81}[/tex] = 0.173
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
Learn more about steam here:
https://brainly.com/question/15447025
#SPJ2
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.
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
The plate is made of steel having a density of 7910 kg/m3 .If the thickness of the plate is 9 mm , determine the horizontal and vertical components of reaction at the pin A and the tension in cable BC.
Answer:
Horizontal component ( Ax) = 0
vertical component ( Ay ) = 1794.87 N
Explanation:
Attached below is the detailed solution and the free body diagram of the question
given data:
density = 7910 kg/m3
thickness of plate = 9 mm = 0.009
determine the horizontal and vertical components of the reaction at the pin A and the tension in the cable BC
In general, MOSFET'S:___________.
A) are mostly used in switching circuits
B) can be fabricated in much higher densities than BJT'S
C) produce simpler circuits than BJTS
D) all of the above
Answer:
A. Are mostly used in switching circuitsExplanation:
MOSFET: The acronym for "metal oxide semiconductor field-effect transistor" are mostly used in switching circuits.
There are two classes of MOSFET
1. Depletion mode
2. Enhancement mode
Generally a MOSFET is a kind of transistor, it is actually a field effect transistor with tree terminals gate, source and drain terminals, also the MOSFET can be used as an amplifier for the amplification of electronic signals in the electronic circuit/devices
A two-lane, one-way ramp from an urban expressway with a design speed of 30 mi/h connects with a local road at a T-intersection. The turning roadway has a vertical curb on both sides. Determine the width of the turning roadway if the predominant turning vehicles are single unit trucks with some semi-trailers. Use 0.08 for super-elevation if applicable.
Answer:
30 feet
Explanation:
Given data :
design speed = 30 miles/h
super elevation = 0.08
determine the width of the turning roadway
calculate the value of R = V^2 / 15( e + p)
e = 0.08 , p = 0.2 , v = 30
R = (30)^2 / 15 ( 0.08 + 0.2 )
= 900 / 15 ( 0.28 )
≈ 215 ft
pavement width from the calculation above = 28 ft
width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )
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
