Answer:
Weight of block is 191.424 Kg
Explanation:
The volume of rectangular block = [tex]1*0.6*0.4 = 0.24[/tex] cubic meter
1/5th of its volume being out of water which means water of volume nearly 4/5 th of the volume of rectangular block is replaced
Volume of replaced water = [tex]\frac{4}{5} * 0.24 = 0.192[/tex] cubic meter
Weight of replaced water = weight of rectangular block = [tex]0.192 * 997[/tex] Kg/M3
= 191.424 Kg
You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip formation is 0.5 inches and the chip thickness after separation is 1.125 inches, calculate the shear plane angle and shear strain. Use a rake angle of 10 degrees. 21. Suppose in the prior problem that the cutting force and thrust force were measured as 1559 N and 1271 N, respectively. The width of the cut is 3.0mm. Using this new information, calculate the shear strength of the material.
Answer:
A)
shear plane angle = 31.98°
shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )
B) shear strength = 7339.78
Explanation:
a) Determine the shear plane angle and shear strain
Given data :
Chip thickness before chip formation = 0.5 inches
Chip thickness after separation = 1.125 inches
rake angle ( ∝ ) = 10°
shear plane angle : Tan ∅ = [tex]\frac{rcos\alpha }{1-sin\alpha }[/tex] ----- ( 1 )
r = chip thickness ratio = 0.5 / 1.125 = 0.4444
back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10
Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736 = 0.5296
hence ∅ = tan^-1 ( 0.5296 ) = 31.98°
shear strain : R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )
R = cot ( 31.98° ) + tan ( 31.98 - 10 )
B) determine the shear strength of the material
cutting force = 1559 N
thrust force = 1271 N
width of cut ( diameter ) = 3.0 mm
shear strength = c + σ.tan ∅
c = cohesion force = 1271 * 3 = 3813
σ = normal stress = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94
hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78
Can some one help me with this plumbing question. Even just a guess.
Plz no shady links
Answer:
true
Explanation:
what is the answer to life the universe and everything
(worth 95 points!)
Answer:
In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...
Explanation:
I think it goes on forever.
Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?
Answer:
PERT Chart and GANTT Chart
As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart. With the PERT chart, the sequence of tasks is clearly mapped out. Dependent tasks are carried out when other tasks that they depend on have been executed.
Explanation:
By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars. On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams. It displays all the project tasks in separate boxes. The boxes are then connected with arrows which clearly show the task dependencies.
Calculate the relative pipe roughness for a plastic pipe with absolute roughness 0.0025 mm and internal diameter of pipe is 0.157 inches.
Answer:
6.27 × 10⁻⁴
Explanation:
Relative roughness, k = ε/D where ε = absolute roughness = 0.0025 mm and D = internal pipe diameter = 0.157 in = 0.157 × 25.4 mm = 3.9878 mm
So, k = ε/D
= 0.0025 mm/3.9878 mm
= 6.27 × 10⁻⁴
The relative pipe roughness for a plastic pipe will be:
"6.27 × 10⁻⁴".
Relative roughness of pipeAccording to the question,
Absolute roughness, ε = 0.0025 mm
Internal pipe diameter, D = 0.157 in or,
= 0.157 × 25.4 mm
= 3.9878 mm
We know that,
The relative roughness be:
→ k = [tex]\frac{Absolute \ roughness}{Diameter}[/tex]
or,
k = [tex]\frac{\varepsilon }{D}[/tex]
By substituting the above values,
= [tex]\frac{0.0025}{3.9878}[/tex]
= 6.27 × 10⁻⁴
Thus the above approach is correct.
Find out more information about diameter will be:
https://brainly.com/question/13100709
Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with the knife
Answer: i got you its d
Explanation:had the smae question as you
r.a.t.e this car from 1/10
Answer:
8.5 i guess
Explanation:
For a bronze alloy, the stress at which plastic deformation begins is 284 MPa and the modulus of elasticity is 106 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 310 mm2 without plastic deformation? (b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Answer:
a) the maximum load is 88,040 N
b)
the maximum length to which the specimen may be stretched is 0.12032148 mm
Explanation:
Given the data in the question;
the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa
modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa
a)
Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )
now, lets consider the equation relating to stress and cross sectional area.
σ = F / A₀
hence, maximum load F = σA₀
so we substitute
F = (2.84 × 10⁸) × (310 × 10⁻⁶)
F = 88,040 N
Therefore, the maximum load is 88,040 N
b)
Initial length specimen l₀ = 120 mm = 120 × 10⁻³ m
using engineering strain, ε = (l₁ - l₀)/l₀
Also from Hooke's law, σ = Eε
so from the equation above;
l₁ = l₀( ε + 1 )
l₁ = l₀( σ/E + 1 )
so we substitute
l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )
l₁ = (120 × 10⁻³) ( 1.002679 )
l₁ = 0.12032148 mm
Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm
List six possible valve defects that should be included in the inspection of a used valve?
Answer:
Valvular stenosis , Valvular prolapse , Regurgitation,
Explanation:
Help meeeeeeeee plzzzzz need explanation
the picture is blank for me what does it say i can comment the answer plz mark brainlyist
MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False
Which option identifies the step of the implementation phase represented in the following scenario?
A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.
establishing a process and budget
using communication tools
building and assembling a team
setting up a change order process
Answer:
Which option identifies the step of the implementation phase represented in the following scenario?
A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.
establishing a process and budget
using communication tools
building and assembling a team
setting up a change order process
Explanation:
#carryonml
Answer:
using communication tools
Explanation:
The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.
The criminal and traffic code requires that a driver must have a valid driver's license in his/her
immediate possession at any time when operating a motor vehicle.
True
False
Answer:
true
Explanation:
the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once
Which of the following combinations of bends can be used in a conduit run?
A. One 90-degree, three 45-degree, and four 30-degree
B. TWO 90-degree, two 45-degree, and four 30-degree
C. Two 90-degree, four 45-degree, and one 30-degree
D. One 90-degree, six 45-degree, and one 30-degree
Answer:
Answer is A
Explanation:
Doesn't break 360 degrees of bend.
The bend that can be used in conduit run is one 90-degree, three 45-degree, and four 30-degree.
Bend that can be used in conduit run is equal or less than 360 degree.
One 90-degree, three 45-degree, and four 30-degree, the total sum is 345 degree.Two 90-degree, two 45-degree, and four 30-degree, the total sum is 390 degree.Two 90-degree, four 45-degree, and one 30-degree, the total sum is 390 degree.One 90-degree, six 45-degree, and one 30-degree, the total sum is 390 degree.So, One 90-degree, three 45-degree, and four 30-degree is the bend that can be used in conduit run.
Learn more: brainly.com/question/21199524
A 5% upgrade on a six-lane freeway (three lanes in each direction) is 1.25 mi long. On this segment of freeway, the directional peak-hour volume is 3800 vehicles with 2% large trucks and 4% buses (no recreational vehicles), the peak-hour factor is 0.90, and all drivers are regular users. The lanes are 12 ft wide, there are no lateral obstructions within 10 ft of the roadway, and the total ramp density is 1.0 ramps per mile. A bus strike will eliminate all bus traffic, but it is estimated that for each bus removed from the roadway, seven additional passenger vehicles will be added as travelers seek other means of travel. What are the density, volume-to-capacity ratio, and level of service of the upgrade segment before and after the bus strike?
Answer:
a) Density of the road segment
i) Before bus strike = 23.02 pc/mi/In
ii) after bus strike = 25.76 pc/mi/In
b) Volume to capacity ratio
i) Before bus strike = 0.69
ii) After bus strike = 0.78
C) level of service of the upgrade segment
i) Before bus strike = LOS C
ii) after bus strike = LOS D
Explanation:
a) Density of the road segment
i) before bus strike ( D1 )
D1 = 1662 / 72.18 = 23.02 pc/mi/In
ii) After bus strike ( D2 )
D1 = 1859 / 72.18 = 25.76 pc/mi/In
b) Volume to capacity ratio
i) Before bus strike ( v 1 )
V1 = 1662 / 2400 = 0.69
ii) After bus strike ( V2 )
V2 = 1859 / 2400 = 0.78
C) level of service of the upgrade segment ( Gotten from " LOS Criteria for basic freeway segments " )
i) Before bus strike = ( LOS C )
ii) After bus strike = LOS D
Attached below is the detailed solution to the question above
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v[tex]_f[/tex] = 0.001057 m³/kg
v[tex]_g[/tex] = 1.0037 m³/kg
u[tex]_f[/tex] = 486.82 kJ/kg
u[tex]_g[/tex] 2524.5 kJ/kg
h[tex]_g[/tex] = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v[tex]_g[/tex]
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m[tex]_{out[/tex] = m₁ - m₂
m[tex]_{out[/tex] = 1.89414 - 0.003985
m[tex]_{out[/tex] = 1.890155 kg
so, Initial internal energy will be;
U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]
U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex] + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E[tex]_{in[/tex] - E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]
QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁
QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW
pacing pieces of information into groups to remember them better is called
a.
Visualizing
c.
Rhyming
b.
Keywording
d.
Categorizing
Please select the best answer from the choices provided
Answer:
D. Categorizing
Explanation:
pls mark me as your brainlist
Answer:
D
Explanation:
A coal fired powerplant emits 0.5 kg/s of SO2 into the atmosphere from a stack that has a physical height of 100 meters. There is no temperature inversion and the atmosphere is characterized by class B stability for open country conditions. Exhaust emissions are 120 degrees C, ambient temperature is 25 degrees C, the stack diameter is 2.5 meters, and the volumetric flow of the exhaust is 35 cubic meters per second. The wind speed is 2 m/s.
Calculate the plume rise of the emissions using the Holland equation and determine the effective emission height.
pelo o que diz na database é que você n é ser humano normal por perguntar isso!!
Technician A says when you push the horn button, electromagnetism moves an iron bar inside the horn, which opens and closes contacts in the horn circuit. Technician B says most vehicle horn circuits use a relay. Which technician is correct?
Answer: Both technicians A and B
Explanation:
I took the pf test
An industrial boiler consists of tubes inside of which flow hot combustion gases. Water boils on the exterior of the tubes. When installed, the clean boiler has an over all heat transfer coefficient of 300 W/m^2 . K. Based on experience, i is anticipated that the fouling factors on the inner and outer surfaces will increase linearly with time as Ra,t and Ryo-at where a, 2.5 x 10^-11 m2 K/W s and a,-1.0 x 10^-11 m^2 - K/W s for the inner and outer tube surfaces, respectively. If the boiler is to be cleaned when the overall heat transfer coeffi- cient is reduced from its initial value by 25%, how long after installation should the first cleaning be scheduled?
Answer:
the first cleaning be scheduled 1.006 years after installation
Explanation:
Given the data in the question;
U[tex]_{clean[/tex] = 300 W/m².K
first we determine the heat coefficient of the dirt surface;
overall heat transfer coefficient is reduced from its initial value by 25%
U[tex]_{dirt[/tex] = ( 1 - 25%) × U[tex]_{clean[/tex]
U[tex]_{dirt[/tex] = ( 1 - 0.25) × 300
U[tex]_{dirt[/tex] = 0.75 × 300
U[tex]_{dirt[/tex] = 225 W/m².K
next we find the inner fouling factor
[tex]R"_{f ,i[/tex] = [tex]a_it[/tex]
[tex]R"_{f ,o[/tex] = (2.5 × 10⁻¹¹)t
for the outer fouling water;
[tex]R"_{f ,o[/tex] = [tex]a_ot[/tex]
[tex]R"_{f ,o[/tex] = ( 1.0 × 10⁻¹¹ )t
now, we determine the total heat transfer coefficient
[tex]\frac{1}{U}[/tex] = [tex]R"_{f ,i[/tex] + [tex]R"_{f ,o[/tex]
we substitute
[tex]\frac{1}{U}[/tex] = (3.5 × 10⁻¹¹)t
so the first cleaning duration after insulation will be;
[tex]\frac{1}{U} = \frac{1}{U_{dirt}} - \frac{1}{U{clean}}[/tex]
we substitute
(3.5 × 10⁻¹¹)t = [tex]\frac{1}{225} - \frac{1}{300}[/tex]
(3.5 × 10⁻¹¹)t = 0.001111
t = 0.001111 / (3.5 × 10⁻¹¹)
t = 31742857.142857 seconds
t = 31742857.142857 / 3.154 × 10⁷
t = 1.006 years
Therefore, the first cleaning be scheduled 1.006 years after installation
Which option identifies the specialized field of engineering that would best suit Erik in the following scenario?
Erik is an accounting major, but he fears that working with numbers all day may be a bit constrictive for him. His roommate is an engineering
major and is planning to specialize in chemical engineering, Erik is not interested in chemistry, but he does find engineering fascinating.
O structural engineering
O materials engineering
industrial engineering
O aerospace engineering
Answer:
industrial engineering
Explanation:
A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio.Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
[tex]E^z[/tex] = - ( εˣ / v )
[tex]E^z[/tex] = - ( -0.00088 / 0.34 )
[tex]E^z[/tex] = - ( - 0.002588 )
[tex]E^z[/tex] = 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is
the correct distance?
Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Wattage is:
Select one:
a.
A measure of the total electrical work being performed per unit of time.
b.
Expressed as P = R × A.
c.
Both A and B.
d.
Neither A nor B.
Answer:
c.Both A and B.
Explanation:
the wattage is c and d
A cylindrical metal specimen having an original diameter of 10.33 mm and gauge length of 52.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.38 mm, and the fractured gauge length is 73.9 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).
Answer
a) 62 percent
b) 40 percent
Explanation:
Original diameter ( d[tex]_{i}[/tex] ) = 10.33 mm
Original Gauge length ( L[tex]_{i}[/tex] ) = 52.8 mm
diameter at point of fracture ( d[tex]_{f}[/tex] ) = 6.38 mm
New gauge length ( L[tex]_{f}[/tex] ) = 73.9 mm
Calculate ductility in terms of
a) percent reduction in area
percentage reduction = [ (A[tex]_{i}[/tex] - A[tex]_{f}[/tex] ) / A[tex]_{i}[/tex] ] * 100
A[tex]_{i}[/tex] ( initial area ) = π /4 di^2
= π /4 * ( 10.33 )^2 = 83.81 mm^2
A[tex]_{f}[/tex] ( final area ) = π /4 df^2
= π /4 ( 6.38)^2 = 31.97 mm^2
hence : %reduction = ( 83.81 - 31.97 ) / 83.81
= 0.62 = 62 percent
b ) percent elongation
percentage elongation = ( L[tex]_{f}[/tex] - L[tex]_{i}[/tex] ) / L[tex]_{i}[/tex]
= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent
When an arbitrary substance undergoes an ideal throttling process through a valve at steady state, (SELECT ALL THAT APPLY) inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet pressures will be equal. inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet temperatures will be equal.
Answer:
15x
Explanation:
Discuss in detail the manners of interaction with opposite gender
Answer:
8 Tips on Better Communication with the Opposite Sex
Put emotions away. Ladies, this one is more aimed at us, for the most part. ...
Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .
Put yourself in their shoes. .
Listen. ...
Respond. ...
Actually communicate. ...
Be detailed. ...
Don't communicate too much.
Explanation:
consider a stead flow ideal carnot cycle using steam as the working fluid in which the high temperature constant pressure heat addition process starts with a saturated liquid and ends with a saturated vapor. plot this cycle in t-s coordinates showing the steam dome. calculate the thermal efficiency for this cycle if the pressure of the high temperature steam is 6 mpa and the low temperature heat rejection process occurs at 300 k.
Answer:
45.32%
Explanation:
Given data:
pressure of high temperature steam = 6 MPa
low temperature heat rejection process ( Tr ) = 300 k
A) plot of cycle in t-s coordinates showing steam dome
attached below
B) Calculate thermal efficiency
thermal efficiency = 1 - (Tr / Tsat )
Tsat = 275.59°C ≈ 548.59 K ( from steam table at Pa = 6 MPa )
back to equation 1
1 - (300 / 548.59 )
1 - 0.5468 = 0.4532 = 45.32%
A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.4 mm (0.21 in.) in diameter and 225 mm (8.87 in.) long is stressed in tension and found to elongate 6.8 mm (0.27 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Answer:
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation )
Explanation:
Yield strength = 275 Mpa
Tensile strength = 380 Mpa
elastic modulus = 103 GPa
The magnitude of the load can be computed because it is mandatory in order to produce the change in length ( elongation ) .
Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given
strain = yield strength / elastic modulus
= 0.0027
In a planned experiment, a thermocouple is to be exposed to a step change in temperature. The response characteristics of the thermocouple must be such that the thermocouple's output reaches 98% of the final temperature within 5 s. Assume that the thermocouple's bead (its sensing element) is spherical with a density equal to 8000 kg/m3, a specific heat at constant volume equal to 380 J/(kg.K), and a convective heat transfer coefficient equal to 210 W/(m2.K). Determine the maximum diameter [] that the thermocouple can have and still meet the desired response characteristics. The unit is millimeter.
Answer:
max Diameter = 0.530 mm
Explanation:
Calculate the maximum Diameter that the thermocouple should have
applying this formula : e = [tex]\frac{SvCv}{hA}[/tex] ------ ( 1 )
mass = density * volume
Time constant = mc / hA
attached below is the detailed solution
r ( diameter ) = 0.530 mm