What shortcoming does a parity bit have when used as an error detection scheme for a sequence of ones and zeros
Answer:
It only detects errors if an odd number of bit flips have occurred.
Explanation:
Parity is a measure of how even or odd a string of digits (for example, a string of binary units is). It is obtained by calculating the value of all the bits. Parity is used in error detection. A parity error is detected when the parity and the odd number of bits are incorrectly propagated.
A problem commonly associated with the parity bits is that they can only detect an odd number of bit errors. This means that if a digit is corrupted in the even binary digits there would be an incorrect detection of errors.
This can change the fit of your respirator A) A mustache B) A beard C) Weight gain D) All of the above
Answer:
A) A mustache
B) A beard
Explanation:
A respirator is a device which is worn over the face to aid breathing through the use of oxygen. Because the fit is won on the face a mustache, or beard would change the fit of the respirator of the face as this would make the respirator tighter in some cases while in others not fit at all.
Answer: D) All of the above!
Explanation:
inspections may be_____ or limited to a specific area such as electrical or plumbing
A. Metering
B. General
A commercial jet is flying at a standard altitude of 35,000 ft with a velocity of 550 mph: (a) what is the Mach number? (b) should the flow be treated as incompressible, why or why not?
Answer:
Mach number = 0.68168
The flow should be treated as compressible.
Explanation:
Given that:
The altitude of a commercial jet = 35000
The properties of air at that given altitude are as follows:
Pressure = 24.577 kPa
Temperature T = 50.78176° C
Temperature T = ( 50.78176 + 273 )K = 328.78176 K
[tex]\varphi = 0.38428 \ kg/m^3[/tex]
The velocity is also given as: 550 mph = 245.872 m/s
Therefore, the sonic velocity is firstly determined by using the formula:
[tex]a = \sqrt{ \vartheta \times R \times T\[/tex]
[tex]a = \sqrt{1.4 \times 287 \times 323.78176[/tex]
[tex]a = \sqrt{130095.5112[/tex]
a = 360.68755 m/s
Then, we can calculate the Mach number by using the expression:
[tex]{Mach \ number = \dfrac{V}{a}}[/tex]
[tex]Mach \ number = \dfrac{245.872}{360.68755}[/tex]
Mach number = 0.68168
b) Ideally, all flows are compressible because the Mach number is greater than 0.3, suppose the Mach number is lesser than 0.3, then it is incompressible.
A 30-mm-diameter shaft, made of AISI 1018 HR steel, transmits 10 kW of power while rotating at 200 rev/min. Assume any bending moments present in the shaft to be negligibly small compared to the torque. Determine the static factor of safety based on:a) The maximum-shear-stress failure theory.b) The distortion-energy failure theory.
Answer:
a) According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.
b) According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.
Explanation:
First, we need to determine the torque experimented by the shaft ([tex]T[/tex]), measured in kilonewton-meters, whose formula is described:
[tex]T = \frac{\dot W}{\omega}[/tex] (Eq. 1)
Where:
[tex]\dot W[/tex] - Power, measured in kilowatts.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
If we know that [tex]\dot W = 10\,kW[/tex] and [tex]\omega = 20.944\,\frac{rad}{s}[/tex], then the torque experimented by the shaft:
[tex]T = \frac{10\,kW}{20.944\,\frac{rad}{s} }[/tex]
[tex]T =0.478\,kN\cdot m[/tex]
Let consider that shaft has a circular form, such that shear stress is determined by the following formula:
[tex]\tau = \frac{16\cdot T}{\pi\cdot D^{3}}[/tex] (Eq. 2)
Where:
[tex]D[/tex] - Diameter of the shaft, measured in meters.
[tex]\tau[/tex] - Torsional shear stress, measured in kilopascals.
If we know that [tex]D = 0.03\,m[/tex] and [tex]T =0.478\,kN\cdot m[/tex], the torsional shear stress is:
[tex]\tau = \frac{16\cdot (0.478\,kN\cdot m)}{\pi\cdot (0.03\,m)^{3}}[/tex]
[tex]\tau \approx 90164.223\,kPa[/tex]
a) According to the maximum-shear-stress failure theory, we get that maximum shear stress limit is:
[tex]S_{ys} = 0.5\cdot S_{ut}[/tex] (Eq. 3)
Where:
[tex]S_{ys}[/tex] - Ultimate shear stress, measured in kilopascals.
[tex]S_{ut}[/tex] - Ultimate tensile stress, measured in kilopascals.
If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:
[tex]S_{ys} = 0.5\cdot (440\times 10^{3}\,kPa)[/tex]
[tex]S_{ys} = 220\times 10^{3}\,kPa[/tex]
Lastly, the static factor of safety of the shaft ([tex]n[/tex]), dimensionless, is:
[tex]n = \frac{S_{ys}}{\tau}[/tex] (Eq. 4)
If we know that [tex]S_{ys} = 220\times 10^{3}\,kPa[/tex] and [tex]\tau \approx 90164.223\,kPa[/tex], the static factor of safety of the shaft is:
[tex]n = \frac{220\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]
[tex]n = 2.440[/tex]
According to the maximum-shear-stress failure theory, the static factor of safety of the shaft is 2.440.
b) According to the distortion-energy failure theory, we get that maximum shear stress limit is:
[tex]S_{ys} = 0.577\cdot S_{ut}[/tex] (Eq. 5)
If we know that [tex]S_{ut} = 440\times 10^{3}\,kPa[/tex], the ultimate shear stress of the material is:
[tex]S_{ys} = 0.577\cdot (440\times 10^{3}\,kPa)[/tex]
[tex]S_{ys} = 253.88\times 10^{3}\,kPa[/tex]
Lastly, the static factor of safety of the shaft is:
[tex]n = \frac{253.88\times 10^{3}\,kPa}{90164.223\,kPa}[/tex]
[tex]n = 2.816[/tex]
According to the distortion-energy failure theory, the static factor of safety of the shaft is 2.816.
what is most delicious fod in the philippines?
Answer:
Well there are a lot of delicious food in the philppines but my most favorite is the Lechon, Adobo, Sisig, Chicken Curry, Crispy pata and Sinigang
A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of dimensions 5.00 cm by 6.11 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf?
Answer:
The field strength needed is 0.625 T
Explanation:
Given;
angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s
area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²
number of tuns of the coil, N = 300 turns
peak emf = 24 V
The peak emf is given by;
emf₀ = NABω
B = (emf₀ ) / (NA ω)
B = (24) / (300 x 0.003055 x 41.893)
B = 0.625 T
Therefore, the field strength needed is 0.625 T
Recently, due to rapid urbanization and mechanization residents of a city are suffering from harmful effects of the ultra violet rays. Depletion of which layer is likely to have led to this situation
Before you calculate the moment capacity for a steel beam, you have to determine the classification of beam.
a. True
b. False
Answer:
True
Explanation:
True - because different classification of steel beam have different yield strength.
The moment capacity for a steel beam is given by;
M = Mn / Ωₙ
where;
M - the maximum moment acting on the beam
Ωₙ - is the Safety Factor for Elements in Bending = 1.67
Mn - nominal moment of the steel, given as
[tex]M_n = Z_x *f_y[/tex]
where;
Zₓ - the Plastic Section Modulus in the x or strong axis.
[tex]f_y -[/tex] is the Yield Strength of the Steel (A36W, A46 W or A50W)
A36W = 36 ksi
A46 W = 46 ksi
A50W = 50 ksi
Thus, before you calculate the moment capacity for a steel beam, you have to determine the classification of beam, for the yield strength of the steel beam.
Which metal is stronger? Tungsten or Titanium?
Answer:
i believe tungsten is stronger than titanium !
what is the discount on airpod pro thank you screen
Answer:
190$-200$
Explanation:
Estimate the rotor inertia assuming that the rotor is a cylinder of radius 8.98 mm, and length 25 mm, with a material of 100% copper. Explain why the rotor inertia may differ from these assumptions?
Answer:
The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.
The rotor inertia may differ from these assumption due to differences in the shape of cross section.
Explanation:
We assume that rotor can be represented as a solid cylinder of radius [tex]r[/tex], length [tex]l[/tex], made of cooper ([tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex]) and whose axis of rotation passes through its center of mass and is parallel to its cross section. By definition of Moment of Inertia and Theorem of Parallel Axes, the moment of inertia of the rotot is:
[tex]I = \frac{1}{4}\cdot \rho \cdot \left(\frac{\pi}{4} \right) \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex]
[tex]I = \frac{\pi}{16}\cdot \rho \cdot R^{3}\cdot (3\cdot R^{2}+L^{2})[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of copper, measured in kilograms per cubic meter.
[tex]R[/tex] - Radius of the rotor, measured in meters.
[tex]L[/tex] - Length of the rotor, measured in meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
If we know that [tex]\rho = 8960\,\frac{kg}{m^{3}}[/tex], [tex]L = 25\times 10^{-3}\,m[/tex] and [tex]R = 8.98\times 10^{-3}\,m[/tex], the estimated moment of inertia of the rotor is:
[tex]I = \frac{\pi}{16}\cdot \left(8960\,\frac{kg}{m^{3}} \right)\cdot (8.98\times 10^{-3}\,m)^{3}\cdot [3\cdot (8.98\times 10^{-3}\,m)^{2}+(25\times 10^{-3}\,m)^{2}][/tex]
[tex]I \approx 1.105\times 10^{-6}\,kg\cdot m^{2}[/tex]
The moment of inertia of the rotor is approximately [tex]1.105\times 10^{-6}[/tex] kilogram-square meters.
From D'Alembert's Formula we know that net force of rigid bodies experimenting rotation equals the product of moment of inertia and angular acceleration. In this case, the purpose is minimizing moment of inertia and it is done by modifying the shape of the cross section so that rotor could be aerodynamically more efficient.
