Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says that a way to prevent galvanic corrosion is to reuse coated bolts. Who is right
Answer:
Technician A
Explanation:
Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.
GMA MIG weiding is a
Gas metal arc welding (GMAW), sometimes referred to by its subtypes metal inert gas (MIG) welding or metal active gas (MAG) welding, is a welding process in which an electric arc forms between a consumable MIG wire electrode and the workpiece metal(s), which heats the workpiece metal(s), causing them to melt and join.
Explanation:
GMA MIG weiding is aiejrjjrkdkff
A good attitude toward blank means believing that the proper attitude and habits are extremely important
Answer:
the blank = life
Explanation:
:) happy holidays!
1. The construction process begins with which of the following stages?
a) establishing the foundation
b) enclosing the building
c) site preparation
d) installing underground utilities
Answer:
c) site preparation
Explanation:
A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.
Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.
This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.
Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.
Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.
Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kept at a temperature of 80.6 °F year round using a natural gas heater. If you were to completely drain the pool and refill the pool with 50°F water, (blank) GJ (giga-Joules) of energy are required to to heat the water back to 80.6 °F. Note: The specific heat capacity of water is 4182 J/kg ⋅°C. The cost of natural gas per GJ is $2.844. It costs $ (blank) to heat the pool (to the nearest dollar).
Answer:
[tex]95.914\ \text{GJ}[/tex]
[tex]\$272.78[/tex]
Explanation:
m = Mass of water = 749511.5 kg
c = Specific heat of water = 4182 J/kg ⋅°C
[tex]\Delta T[/tex] = Change in temperature = [tex]80.6-50=30.6^{\circ}\text{F}[/tex]
Cost of 1 GJ of energy = $2.844
Heat required is given by
[tex]Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}[/tex]
Amount of heat required to heat the water is [tex]95.914\ \text{GJ}[/tex].
Cost of heating the water is
[tex]95.914\times 2.844=\$272.78[/tex]
Cost of heating the water to the required temperature is [tex]\$272.78[/tex].
On a Test please help
When towing a trailer, you should consider all of the following except
the minimum speed limit on roadways.
your personal protective equipment.
the gross weight of the load.
the increased stopping distance
Answer:
I'm not 100%sure but I think it's the first one
Compare and contrast the roles of agricultural and environmental scientists.
Answer:
HUHHHHHH BE SPECIFIC CHILE
Explanation:
ERM IRDK SORRY BOUT THAT
A garden hose fills a 2-gallon bucket in 5 seconds. The number of gallons, g (y), is proportional to the number of seconds, t (x), that the water is running. What is the constant of proportionality?
Answer:
0.4 gallons per second
Explanation:
A function shows the relationship between an independent variable and a dependent variable.
The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.
The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).
Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.
Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second
All of these are part of the seat belt assembly EXCEPT the:
O latch plate
O D-ring
O retractor.
O cushion
A rectangular bar has a edge crack at the bottom and is subjected to a pure bending moment. The crack length is a = 1 mm. The height of the bar is b = 12.5 cm. Knowing that the failure strength of the material is Sigma = 1,400 MPa, what is the fracture toughness of the material, K_ic.
Answer:
The answer is "[tex]\bold{87.3906 \ MPa \sqrt{m}}[/tex]".
Explanation:
Given value:
[tex]\sigma = 1400 \ MPa \ \ \ \ \ \ \ \ where \ \sigma = failure \ strength\\\\a = 1 \ mm = 1 \times 10^{-3} \ m \ \ \ \ \ \ \ \ \ \ where\ a = crack\ length\\\\b= 12.5 \ cm = 125 \ mm = 0.125 \ m\\\\[/tex]
[tex]\to \alpha = \frac{a}{b} =\frac{1}{125} = 8 \times 10^{-3}\\\\[/tex]
[tex]k_{b} = \frac{1.12 + \alpha (2.62 \alpha -1.59)}{1-0.7 \alpha}\\[/tex]
[tex]= \frac{1.12 + (8\times 10^{-3}(2.62(8\times 10^{-3}) -1.59))}{1-(0.7 \times 8\times 10^{-3})}\\\\= \frac{1.12 + (8\times 10^{-3}(0.02096 -1.59))}{1-(0.7 \times 8\times 10^{-3})}\\\\= \frac{1.12 + (8\times 10^{-3}(-1.56904))}{1-(0.0056)}\\\\= \frac{1.12 + (-0.01255232)}{0.9944}\\\\= \frac{-1.10744768}{0.9944}\\\\= -1.11368431\\\\[/tex]
[tex]k_{ic} = \sigma \sqrt{\pi a} \ y_b[/tex]
[tex]=1400 \times \sqrt{\pi \times 1 \times 10^{-3} } \times -1.11368431\\\\=1400 \times 0.00177200451 \times -1.11368431\\\\=87.3906 \ MPa \sqrt{m}[/tex]
sin x +√3 coax= √2
Answer:
The two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12
Explanation:
The given equation is
Sin x +√3 Cosx= √2
Upon dividing the equation by 2 we get
[tex]\frac{1}{2}Sinx + \frac{\sqrt{3} }{2}Cosx = \frac{\sqrt{2} }{2}[/tex]
Sin([tex]\frac{pi}{6}[/tex])*Sinx + Cos([tex]\frac{pi}{6}[/tex])*Cosx = [tex]\frac{1}{\sqrt{2} }[/tex]
This makes the formula of
CosACosB + SinASinB = Cos(A-B)
Cos(x-[tex]\frac{pi}{6}[/tex]) = [tex]\frac{1}{\sqrt{2} }[/tex]
cos(x- pi/6) = cos(pi/4)
upon writing the general equation we get
x-pi/6 = 2n*pi ± pi/4
x = 2n*pi ± pi/4 -pi/6
so we will have two solutions
x = 2n*pi + pi/4 -pi/6
= 2n*pi + pi/12
and
x = 2n*pi - pi/4 -pi/6
= 2n*pi -5pi/12
Therefore the two values of x are 2n*pi + pi/12 and 2n*pi -5pi/12.
In urban area you can except
Answer:
eoidnfoejsdncodsnc
Explanation:
dfdjsncojnsdjcnsdojnvjsdvkjsdkjvnsdjvnskjnvkjdvkjnsdcjndkjndskjndskjndskjnsdvkjnvsdkjvsdkjnvskdjnvkjdsvkjsdnvkjsdnkjsvnkj
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R14B, (6) G1B, (7) M8C3, (8) P1NSE, and (9) RL1.
(a) Determine the normal times in TMUs for these motion elements.
(b) What is the total time for this work element in sec
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
A trapezoidal section has a 5.0-ft bed width, 2.5-ft depth, and 1:1 side slope. Evaluate its geometric elements (Area, water depth, wetted perimeter, top width, hydraulic radius, and hydraulic depth).
Answer:
a) 18.75 ft^2
b) 2.5 ft
c) 12.07 ft
d) 10 ft
e) 1.553 ft
f) 1.875 ft
Explanation:
Given data :
5.0-ft bed width, ( b )
2.5-ft depth ( y )
1 : 1 side slope
Evaluate
a) Area of trapezoidal section
A = by + my^2
we assume m = 1
A = [5 + (1 * 2.5 ) ] *2.5
= ( 7.5 ) 2.5 = 18.75 ft^2
b) Calculate water depth
water depth = 2.5 ft
c) Calculate wetted perimeter
P = b + 2y √ 1 + m^2
= 5 + (2.5*2) √ 1 + 1 ^2 = 12.07 ft
d) calculate top width
T = b + 2my
= 5 + 2 ( 1 * 2.5 ) = 10 ft
e) calculate hydraulic radius
R = A / P = 18.75 / 12.07
= 1.553 ft
f) calculate hydraulic depth
D = A / T = 18.75 / 10 = 1.875 ft
If 1 inch equals 10 feet, what would the measured distance be if the line scaled 1 1/2 inches?
A. 10 feet
b. 15 feet
c. 2 feet
d. 25 feet
Answer: B. 15 feet
Explanation:
the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate
This question is incomplete, the complete question is;
the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate ( in m³ ) if the rectangular channel is 20 m wide.
Answer:
the flow rate is 32.549 m³/sec
Explanation:
Given that
y₁ = 0.3 m
y₂ = 1.2 m
β = 20 m
Now for Rectangular Channel, we know that;
2q²/g = y₁y₂( y₁ + y₂)
where g = 9.81 m/s²
and q = Q/β
so
2(Q/β)²/g = y₁y₂( y₁ + y₂)
we substitute our given values
2(Q/20)²/9.81 = 0.3 × 1.2( 0.3 + 1.2)
2(Q²/400)/9.81 = 0.36(1.5)
2(Q²/400) = 0.54 × 9.81
Q²/400 = 5.2974 / 2
Q²/400 = 2.6587
Q² = 1059.48
Q = √1059.48
Q = 32.549 m³/sec
Therefore the flow rate is 32.549 m³/sec
A site is underlain by a soil that has a unit weight of 118 lb/ft3. From laboratory shear strength tests that closely simulated the field conditions, the total stress parameters were measured to be C total = 250 lb/ft2 and φ total = 29°. Estimate the shear strength on a horizontal plane at a depth of 12 ft below the ground surface at this site in lbs/ sq ft
Answer: the shear strength at a depth of 12 ft is 1034.9015 lb/ft²
Explanation:
Given that;
Weight of soil r = 118 lb/ft³
stress parameter C = 250 lb/ft²
φ total = 29°
depth Z = 12 ft
The shear strength on a horizontal plane at a depth of 12ft
ζ = C + δtanφ
where δ = normal stress
normal stress δ = r × z = 118 × 12 = 1416
so
ζ = C + δtanφ
ζ = 250 + 1416(tan29°)
ζ = 250 + 1416(tan29°)
ζ = 250 + 784.9016
ζ = 1034.9015 lb/ft²
Therefore the shear strength at a depth of 12 ft is 1034.9015 lb/ft²
An ECG has a scalar magnitude of 1 mV on lead II and a scalar magnitude of 0.5 mV on lead III. Calculate the scalar magnitude on lead I.
Answer: the scalar magnitude on lead I is 0.5 mV
Explanation:
Given that;
scalar magnitude on lead II = 1 mV
scalar magnitude on lead III = 0.5 mV
the scalar magnitude on lead I = ?
we know that;
Lead I Voltage = LA - RA -----------let this be equation 1
where LA is left arm electrode and RA is right am electrode
Also
Lead II = LL - RA
where LL is the left leg of electrode
we substitute
1 mV = LL - RA ---------------------let this be equation 2
Again
Lead III = LL - LA
we substitute
0.5 mV = LL - LA ------------------let this be equation 3
now subtract equation 3 and 2
1 mV - 0.5 mv = LL - RA - (LL - LA)
0.5 mV = LL - RA - LL + LA
0.5 mV = -RA + LA
0.5 mV = LA - RA
now taking a look at our equation 1 ( Lead I Voltage = LA - RA )
hence, Lead I Voltage = LA - RA = 0.5 mV
Therefore the scalar magnitude on lead I is 0.5 mV
1) I love to swim. 2) A few years ago, my new year's resolution was to become a faster swimmer. 3) First, I started eating better to improve my overall health. 4) Then, I created a training program and started swimming five days a week. 5) I went to the pool at my local gym. 6) To measure my improvement, I tried to count my laps as I was swimming, but I always got distracted and lost track! 7) It made it very hard for me to know if I was getting faster. 8) This is a common experience for swimmers everywhere. 9) We need a wearable device to count laps, calories burned, and other real-time data. Summarey of the story
5.5.1 Dual-edge detector A dual-edge detector is similar to a rising-edge detector except that the output is asserted for one clock cycle when the input changes from 0 to 1 (i.e., rising edge) and 1 to 0 (i.e falling edge). 1. Design a circuit based on the Moore machine and draw the state diagram and ASM chart. 2. Derive the HDL code based on the state diagram of the ASM chart. 3. Derive a testbench and use simulation to verify operation of the code.
The resistance of a 1,000-foot length of #6 AWG wire at a temperature of 25 degrees C is 0.4028 ohm. Thus, option B is correct.
What is prototype?It is a term used in a variety of contexts, including semantics, design, electronics, and software programming.A prototype is an early sample, model, or release of a product built to test a concept or process or to act as a thing to be replicated or learned from.
The statement that defines a prototype is a prototype is a working model of the proposed system. Say for instance a programmer creates a flowchart or an algorithm of an actual program. The flowchart or the algorithm is the prototype of the actual program. This in other words means that prototypes are models of a system. Hence, the statement that defines a prototype.
Therefore, The resistance of a 1,000-foot length of #6 AWG wire at a temperature of 25 degrees C is 0.4028 ohm. Thus, option B is correct.
Learn more about prototype on:
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A brine solution is 26% salt with 70.0 kg of water evaporated per hour. To produce 195 kg of pure salt (0% moisture) per day, how long should the process operate each day and how much brine must be fed to the evaporator per hour?
The process should operate each day for ___ hours.
The amount of brine that must be fed to the evaporator is ___ kg/h.
Answer:
- the process should operate each day for 7.9286 hours
- the amount of brine that must be fed to the evaporator is 94.594 kg/hr
Explanation:
Given that;
concentration of brine = 26%
so water concentration will be 100% - 26% = 74%
for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)
so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr
Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr
Now for production of 195kg of pure dry salt, number of hours required will be
T = 195 / 24.5945 = 7.9286 hrs
Therefore the process should operate each day for 7.9286 hours.
Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg
Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr
Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr
A commercial facility has a three-phase, 277 volt, 1200 Amp service. What is the required nominal transformer size needed for this building?
a. 750 kVA
b. 500 kVA
c. 575 kVA
d. 633 kVA
e. 1,000 kVA
Answer:
The answer is "Option e".
Explanation:
Given value:
[tex]\to v_{ph}=277 \ \ \ \ \ \ \ I_{ph} =1200\\\\[/tex]
[tex]vA= v_{Ph} I_{ph} \ \ \ \to for \ single \ phase\\\\=3v_{ph} I_{ph} \ \ \ \ \ \to for \ 3 - \phi \\\\s = 3\times 277 \times 1200 \\\\[/tex]
[tex]=997.200 \ K VA\\\\= 1000 \ k VA[/tex]
What does Enter key do?
You cannot click Enter key to start a line if your current is blank?
This is spot to do today
Answer:
See below
Explanation:
Enter-key also called the "Return key," it is the keyboard key that is pressed to signal the computer to input the line of data or the command that has just been typed.It Was the Return KeyThe Enter key was originally the "Return key" on a typewriter, which caused the carriage to return to the beginning of the next line on the paper. In a word processing or text editing application, pressing Enter ends a paragraph. A character code for return/end-of-line, which is different in Windows than it is in the Mac, Linux or Unix, is inserted into the text at that point.
Answer:
True
Explanation:
Once there are two yellow lines having inner broken lines on the two sides of a center traffic lane, what this is trying to tell you is that you can use those lanes to start a left hand turn, or a U-turn from the both directions of traffic. However you cannot use it for passing. This is sometimes misunderstood by road users and drivers.
lara sees her colleague taking a bribe from a customer. what should she do?
A. confront her colleague with a warning
B. tell her other colleagues and spread the news
C. ask for a stake in the bribe
D. gather evidence and approach a trusted supervisor
Answer:
D. Gather evidence an approach a trusted supervisor
Explanation:
B and C are obviously not the answer, A will probably not be useful since the colleague will most likely continue what they are doing
If the length of the arm were to increase, how would the required torque change? a. Required torque would increase b. Required torque would decrease c. Required torque would remain the same d. It is impossible to calculate required torque with the values given.
A _________ is interesting only if the statistics computed from transactions covered by the rule are different than those computed from transactions not covered by the rule.
Answer: Quantitative association rule.
Explanation:
The quantitative rules of association apply to the basic type of rules of association which exists as X and Y, with X and Y consisting of a collection of numerical and/or categorical attributes. Unlike general association laws, where both the left and right sides of the law should have categorical (nominal or discrete) attributes, a numerical attribute must be included in at least one attribute of the quantitative association rule (left or right).
Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has (1) a simple cubic crystal structure, (2) an atomic radius of 0.154 nm, and (3) a saturation flux density of 0.83 tesla. Determine the number of Bohr magnetons per atom for this material. A: What is the volume, in m3, for this unit cell?B: How many atoms are there per m3 in this material?C: What is the number of Bohr magentons per atom for this material?
Answer:
a. 2.9218x10^(-29) m^3
b. 7.1230x10^(28) atoms/m^3
c. 2.0812 BM/atom
Explanation:
Atomic radius r = 0.154 nm
saturation flux density Bs = 0.83 tesla
- the formula for the volume of the simple cubic crystal (Vc) = a^3 = (2r)^3
= (2 x 0.154x10^-9)^3
= 2.9218x10^-29 m^3
- The formula for the Bohr magneton per atom with respect to VC, Bs, permeability of the vacuum Uo and magnetic moment per Bohr magneton Ub is;
Bohr magneton per atom nb = (Bs x VC) / (Ub x Uo)
= (0.83 x 2.9218x10^-29) / (9.27x10^-24) x(1.257x10^-6)
= 2.4251x10^-29 / 1.1652x10^-29
= 2.0812 BM/atom
- Number of atoms per Vc, N = nb / Vc
= 2.0812 / 2.9218x10^-29 = 7.1230x10^28 atoms
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distance that a 1 (GHz) wave travels in graphite such that its field intensity is reduced by 30 (dB).
Answer:
the answer is below
Explanation:
a) The conductivity of graphite (σ) is calculated using the formula:
[tex]\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}[/tex]
where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012
Substituting:
[tex]\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m[/tex]
b) f = 1 GHz = 10⁹ Hz.
[tex]\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm[/tex]
While examining the color of an engine's coolant: Technician A says that because it is orange, the cooling system should be flushed and new coolant put into the system. Technician B says that if the coolant looks rusty, the cooling system should be flushed and new coolant put into the system. Who is correct?
Answer:
Technician B is correct.
Explanation:
In every vehicles, coolants are very important as it keeps the engine cool and keep it running. The coolant serves the purpose of dissipating away the heat that is generated when the engine runs.
When the engine runs, the coolant passes all over the heating surfaces in the engine thus carrying away the heat along with it. With time the color of the coolant changes to rusty brown color when the coolant serves its purpose . It also washes away all the dirt particles from the engine part along with it. Thus the color changes and it is now can be flushed from the engine sump and fresh coolant can be put in to the engine.
Thus in the context, technician B is correct.
who is the strongest avenger i say hulk but who knows at this point
Answer:
or is the strongest evenger she hulk
Explanation:
?????????
Answer:
Thor!
Explanation:
In Thor: Ragnarok he beat the Hulk in order for Hulk to win thor had to be electrocuted and in Avengers: Endgame Thor is seen holding open the "Floodgates" and withstanding the radiation from a dying star, also the fact that Thor is a god means that he is all powerful and the rightful heir to the throne to Asgard, plus the fact that he has defeated Loki multiple times a feat that not even the Hulk has done.