Answer: answer provided in the explanation section.
Explanation:
Weather phenomenons that would impart Aviation Operations in Santa Barbara -
1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.
2. The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.
3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.
4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.
This and more are some factors to look into when considering wheather conditions that would affect aviation operations.
I hope this was a bit helpful. cheers
(A) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomicradius R.
(B) Compute and compare planar density values for these same two planes for molybdenum (Mo).
Answer:
Explanation:
BCC structure can also be called Body-Centred Cubic. In BCC Structure, single atom is located at every corner of the cube, and one is also located at the centre of the structure.
Each of the four cube corners through which this (110) plane travels has an atom, which it shares with four neighboring unit The percentage of the entire crystallographic plane area that is occupied by atoms is known as planar density.
What is planar density ?The packing density of crystals is measured by planar density. A face-centered cubic unit cell's planar density may be determined by following a few straightforward procedures. Determine how many atoms are centered on a certain plane. An FCC crystal, for instance, has 2 atoms on a (1 1 0) plane.
The amount of atoms per unit area on an interest plane defines the planar density, a crucial aspect of a crystal structure.
The percentage of the entire crystallographic plane area that is occupied by atoms is known as planar density. Atomic packing factor analogs in one and two dimensions are linear and planar densities.
Thus, The center atom, however, is totally contained within the unit cell. This means that this BCC (110) plane has an equivalent of 2 atoms.
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Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air resistance, which of the following statements are true when the two balls hit the ground?
(a) The kinetic energy of A is the same as the kinetic energy of B
(b) The kinetic energy of A is half the kinetic energy of B.
(c) The kinetic energy of A is twice the kinetic energy of B.
(d) The kinetic energy of A is four times the kinetic energy of B Explain your answer why.
What is the definition of General Plan Motion? What would be the effective methodology or approach to solve a rigid body kinematics problem?
Answer:
The kinetic energy of A is twice the kinetic energy of B
Explanation:
The true statement when the two balls hit the ground is, the kinetic energy of A is twice the kinetic energy of B. The correct option is (c).
What is kinetic energy?Kinetic energy is the energy of motion, which can be seen as an item or subatomic particle moving. Kinetic energy exists in every moving object and particle.
Kinetic energy is demonstrated by a person walking, a soaring baseball, a crumb falling from a table, and a charged particle in an electric field.
The definition of a General Plan of Motion is every point on the body has a different path. As a result, we must relate the forces to the acceleration of the body's center of mass, as well as the moments to the angular accelerations.
Therefore, the correct option is (c), The kinetic energy of A is twice the kinetic energy of B.
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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.
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.
Which of the following sentences is
punctuated incorrectly?
Answer:
Um is there suppose to be a picture with the sentences on
Answer:
It's B I did the test and I was right on B
Explanation:
I don't know I knew it was B
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.
Consider atmospheric air at 25 C and a velocity of 25 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 125 C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 105 , and 106 .
Answer:
Explanation:
Temperature of atmospheric air To = 25°C = 298 K
Free stream velocity of air Vo = 25 m/s
Length and width of plate = 1m
Temperature of plate Tp = 125°C = 398 K
We know for air, Prandtl number Pr = 1
And for air, thermal conductivity K = 24.1×10?³ W/mK
Here, charectorestic dimension D = 1m
Given value of Reynolds number Re = 105
For laminar boundary layer flow over flat plate
= 3.402
Therefore, hx = 0.08199 W/m²K
So, heat transfer rate q = hx×A×(Tp – To)
= 0.08199×1×(398 – 298)
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
According to the building-up principle or aufbau principle, which subshell is typically filled next after the 3d subshell? A) 4p B) 5f C) 6p D) 5d E) 3p
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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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
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of line has resistance R. For a power line that supplies power to 10,000 households, we can conclude that:________
Answer:
IV > [tex]I^{2} R[/tex]
Explanation:
The current in the power line = I
The voltage in the power line = V
The resistance of the power line = R
Power supplied from the power house = P
power delivered to the households = [tex]p[/tex]
We know that the power supplied to a power line system is proportional to
P = IV ....1
we also know that according to Ohm's law, the relationship between the voltage, resistance, and current through an electrical system is given as
V = IR ....2
substituting equation 2 into equation 1, the power delivered to the households is proportional to the square of the current.
[tex]p[/tex] = [tex]I^{2} R[/tex] ....3
The problem is that when power is delivered across a transmission line, some of the power is loss due to Joules heating effect of the power lines. This energy and power loss is proportional to [tex]I^{2}[/tex] therefore, the electrical power delivered to the households will be less than the electrical power supplied from the power station. This means that
P > [tex]p[/tex]
equating these two powers from equations 1 and equation 3, we have
IV > [tex]I^{2} R[/tex]
When choosing building-construction materials, what kinds of materials would you choose, all other things being equal?
The building-construction materials are materials with a relatively large bulk modulus, would the choose of the constructor. Thus, option (e) is correct.
What is building?The term building refers to the large structure of the bricks. The building was the large structure made by the engineer and the architecture. There was the building are the people are the stay. The building was the used in the permanent basis. There are the different types of the building.
According to the building-construction materials are the engineers to the order of the bulk in the material. The bulk level of the material is the high discount rate of the order. The large amount of the material to the used in the building-construction, that are the order in the bulk of the production.
Therefore, option (e) is correct.
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Your question is incomplete, but most probably the full question was.
either materials with a large or a small bulk modulus.it doesn't matter as long as the building is not too tall.materials with a relatively small shear modulus.materials with a relatively small bulk modulus.materials with a relatively large bulk modulus.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).
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
1. Two types of superchargers include
A. turbine and piston.
B. roots-type and centrifugal.
C. double- and single-acting
D. rotary and reciprocating.
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
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
1. In a mechanical design, it is recommended to use standard size/dimension to overcome uncertainties in stress or material strength.
A. True
B. False
2. Steels' surface fatigue strength can be correlated to their Brinell hardnesses.
A. True
B. False
3. If a material point has three principal stresses 01 = 02 = 03 = 300kpsi, then the maximum shear stress is Tmax = 150kpsi.
A. True
B. False
4. The endurance limit of a material is independent on the size of a part.
A. True
B. False
5. The tensile stress in a bolt is usually calculated through tensile force divided by area of unthreaded rod's cross-section area.
A. True
B. False
6. Edge shear-out or tear-out is one failure mode in bolted or riveted joints.
A. True
B. False
7. Pretension of bolts can be controlled by a torque wrench.
A. True
B. False
8. The material ultimate tensile strength of a spring coil is dependent on its coil diameter.
A. True
B. False
9. Adhesive joints are much stronger in tensile loading than shear loading.
A. True
B. False
10. The dominated loading condition in a helical spring is torsion along coils.
A. True
B. False
11. The principal advantage of conical springs is that the its spring rate is non-linear.
A. True
B. False
12. The dimensions of a bearing is coded by two-digit series numbers according to ABMA, one digit for width and the other one for diameter.
A. True
B. False
13. Three key things to determine selection of bearing are load, life and reliability.
A. True
B. False
14. Straight roller bearings can carry large radial loads and small amount of axial load.
A. True
B. False
15. In the AGMA method of gear analysis, gear strength values are the true material strengths.
A. True
B. False
16. A planetary gear train allows for two degrees of freedom.
A. True
B. False
Answer:
1. A.True
2. B.False
3. B.False
4. A.True
5. A.True
6. A.True
7. A.True
8. B.False
9. B.False
10. A.True
11. B.False
12. A.True
13. A.True
14. B.False
15. B.False
16. A.True
Explanation:
Most steels have endurance and fatigue limit about half the Tensile strength. Tensile strength of material is the limit of stress at which material then breaks. The tensile stress in a bolt is calculated through tensile force.
What's the resistance in a circuit that has a voltage of 60 V and a current of 2 A? A. 10 Ω B. 60 Ω C. 120 Ω D. 30 Ω
Answer:
D. Resistance = 30 ohms
Explanation:
Using Ohm's law
V = I times R
Given:
V = 60 V
I = 2 A
Resistance = V / I = 60 V / 20 A
Resistance = 30 ohms
A fixed mass of saturated water vapor at 400 kpa is isothermally cooled until it is a saturated liquid. Calculate the amount of heat rejected during this process in kJ/kg.
The amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.
What is heat?
Heat is defined as the process of kinetic energy being transferred from an energy source to a medium or an item. These three mechanisms radiation, conduction, and convection can all transmit energy. These three modes of heat transfer are typically combined, and it often happens on its own. The heat energy can be transformed into other types of energy, such as mechanical and kinetic energy. Thermal (heat) energy is a type of energy that is produced by the movement of particles and is a component of all matter.
The process rejects a certain amount of heat, which is
-q = w + Δu
q = - w - Δu
q = - 185.5 kJ/kg - 1948.9 kJ/kg
q = 2133.4 kJ/kg
Thus, the amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.
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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.
Match the following light concepts
- Light is bounced back at same angle
- Light penetrates medium at different angle due to different material densities
- Light bounces at different angles in periodic grid
- Light enters medium at different angles through a grid
- Light EMF field looses one axis component
A Polarized filter
B. Transmission Diffraction
C. Classical Reflection
D. Refraction
E. Reflected Diffraction
Answer:
- Light is bounced back at same angle (Classical Reflection)
- Light penetrates medium at different angle due to different material densities (Refraction)
Light bounces at different angles in periodic grid (Reflected Diffraction)
Light enters medium at different angles through a grid (Transmission Diffraction)
- Light EMF field looses one axis component (Polarized filter)
Explanation:
Reflection is a phenomenon in which waves (light included) bounce back from an obstacle at the same angle of incidence
Refraction is the change in the angle of a wave as it enters the interface of two media. The change in angle is due to the difference in the densities of the two media.
Reflected diffraction occurs when an optical component with a periodic grid, splits, and diffracts light into several beams travelling in different directions. The light light bounces at an angle in the periodic grid.
Transmission diffraction is dispersion a beam of various wavelengths into a spectrum of associated lines due to the principle of diffraction. In this type of diffraction, light enters medium at different angles through a grid.
Polarized filters removes one field from the incidence electromagnetic wave like light, leaving it to vibrate in only one plane.
In order to enhance net radiation exchange between a high temperature surface and a low temperature surface in an enclosure, one should:________.
(A) polish both surfaces so that they are highly reflective
(B) paint both surfaces so that they are close to a blackbody surface
Answer: Option B
Explanation:
In order to enhance net radiation exchange between a high temperature surface and a low temperature surface in an enclosure, one should: paint both surfaces so that they are close to a blackbody surface but don't polish both surfaces so that they are highly reflective
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
Consider the thermocouple and convection conditions of Example 1, but now allow for radiation exchange with the walls of a duct that encloses the gas stream. If the duct walls are at 400℃ and the emissivity of the thermocouple bead is 0.9, calculate the steady-state temperature of the junction
Answer:
hello your question has some missing part attached below is the complete question
answer : steady state temperature = 419.713k ≈ 218.7⁰c
Time required to reach a junction ≈ 5 secs
Explanation:
The detailed solution of the given problem is attached below but the solution to the subsequent problem from which the question you asked is referenced to( problem 1 ), is not attached because it was not part of the question you asked
/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a
mean velocity of 7 m/s, and is cooled by the water outside. If the average heat
transfer coefficient is 85 W/m2
°C and the tube temperature is nearly equal to the
water temperature of 5°C, determine the exit temperature of air and the rate of heat
transfer.
Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
The phase sequence of a 3-phase system for which VAN = 120 /90o V and VBN = 120 /210o V is:_______
a) bca
b) abc
c) cab
d) acb
Answer:
b. abc
Explanation:
The phase rotation and its sequence is order in which voltage is reached to their respective sources. The wave forms of a polyphase flow through AC source to the panel. The three phase system has two possible sequence which can be a-b-c or c-b-a. The correct answer is therefore b.
An ideal Diesel cycle has a compression ratio of 17 and a cutoff ratio of 1.3. Determine the maximum temperature of the air and the rate of heat addition to this cycle when it produces 140 kW of power and the state of the air at the beginning of the compression is 90 kPa and 578C. Use constant specific heats at room temperature.
Answer:
maximum temperature = 1322 k
rate of heat addition = 212 kw
Explanation:
compression ratio = 17
cut off ratio = 1.3
power produced = 140 Kw
state of air at the beginning of the compression = 90 kPa and 578 c
Determine the maximum temperature of air
attached below is the detailed solution
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
A power screw is 30 mm in diameter and has a thread pitch of 5 mm. Find the thread depth, the thread width, the mean and root diameters, and the lead, provided that square threads are used. Assume single threads.
Answer:
thread depth = 2.5 mm
thread width = 2.5 mm
mean diameter = 27.5 mm
root diameter = 25 mm
lead of screw = 5 mm
Explanation:
given data
power screw diameter D = 30 mm
thread pitch P = 5 mm
solution
First, we get here thread depth fr square thread
thread depth = [tex]\frac{P}{2}[/tex] ......................1
thread depth = [tex]\frac{5}{2}[/tex]
thread depth = 2.5 mm
and
thread width for square thread
thread width = [tex]\frac{P}{2}[/tex] ......................2
thread width = [tex]\frac{5}{2}[/tex]
thread width = 2.5 mm
and
mean diameter is
mean diameter = D - [tex]\frac{P}{2}[/tex] ................3
mean diameter = 30 - [tex]\frac{5}{2}[/tex]
mean diameter = 27.5 mm
and
root diameter is
root diameter = D - P ....................4
root diameter = 30 - 5
root diameter = 25 mm
and
lead of screw for single thread so n = 1
so lead of screw = 1 × 5
lead of screw = 5 mm
