Answer:
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Explanation:
The given values are,
σ=1.65 MPa
γs=0.60 J/m2
E= 2.0 GPa
The maximum possible length is calculated as:
[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Vince is trying to figure out the volume of two mystery matters. The volume of one of the substances needs to be measured by submerging it in water and the other needs to be measured using a graduated cylinder. Based on the properties of two mystery matters, what are they? (2 points)
Group of answer choices
A rock and orange juice
Helium and a golf ball
Lemonade and milk
Orange juice and helium
Answer:
Rock and orange juice
Explanation:
The mystery matter to be submerged in water must be a solid, therefore we can eliminate the Lemonade and Milk, and Orange juice and Helium, as these pairs do not contain solids. The graduated cylinder is used to measure the volume of a liquid, therefore the only remaining option is Rock and Orange Juice.
A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction
Answer:
26.02 ft
86.7690 ft/min
Explanation:
After 3 steps
0.75³(2.0 thickness)
T = 0.84375
W = (1+0.03)³10
= 10.92727 inches
A To get length
2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf
= 2880 = 9.21988Lf
Lf = 2880/9.21988
= 312.368 inches
Convert to feet
322.368 x 0.0833
= 26.02 ft
B.
= 2 x 10 x 40 = 0.84375 x 10.92727 x vf
800 = 9.21988vf
Vf = 800/9.21988
Vf = 86.7690 ft/min
The allowable tensile stress for a 6.25 mm diameter bolt with a thread length of 5.5 mm is 207 MPa. The allowable shear stress of the material is 103 MPa. Where and how will such a bolt be most likely to fail if placed in tension
Answer:
At the threads due to shear.
Explanation:
Given :
The allowable tensile stress = 207 MPa
The allowable shear stress = 103 MPa
If a tensile force is applied, the maximum shear stress occurs at the threads of the bolt. The bolt is most likely to fail at the critical section. The critical cross section is the section having the minimum cross sectional area.
The portion of the bolt having threads has the minimum cross sectional area.
So when the bolt is applied with a tensile force, failure is most likely to take place at the threads due to the shearing force.
Hi, can anyone draw me an isometric image of this shape?
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Answer:do me ti
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Explanation:
mitski
Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 oF and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by Cp
Complete Question:
Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by
C_p=10.03+0.0184T C_p[=]Btu/lbmole- "F ; T[=] °F C,
Answer:
[tex]P'=0.377hp[/tex]
Explanation:
From the question we are told that:
Initial Temperature T_1=70 F
Final Temperature [tex]T_2=250pisa =114.94F[/tex]
Efficiency [tex]E=75\%=0.75[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=\int C_pT[/tex]
[tex]W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )[/tex]
[tex]W=527.21btu/ibmole[/tex]
[tex]W=11.982btu/ibm[/tex]
Generally the equation for Efficiency is mathematically given by
[tex]E=\frac{isotropic Power}{Actual P'}[/tex]
[tex]E=\frac{P}{P'}[/tex]
Since
Isotropic Power
[tex]P=0.0167*11.982btu/ibm[/tex]
[tex]P=0.2btu/s[/tex]
Therefore
[tex]P'=\frac{0.2}{0.75}[/tex]
[tex]P'=0266btu/s[/tex]
Since
[tex]1btu/s=1.4148hp[/tex]
Therefore
[tex]P'=0.377hp[/tex]
to check for ripple voltage from the alternator, connect a digital multimeter and select
Answer:
isn't it summer? sjsushsiansudndd
An aggregate blend consists of 65% of aggregate A and 35% of aggregate B. The bulk specific gravities of aggregate A and B are 2.45 and 3.25, respectively. What is the bulk specific gravity of the blend?
a) 2.45
b) 2.68
c) 2.73
d) 2.92
Answer:
2.68
Explanation:
Percentage by Mass of each Aggregate :
Pa = 65% ; Pb = 35%;
Bulk Specific gravity of each aggregate :
Ga = 2.45 ; Gb = 3.25
Gsb = (Pa + Pb) / (Pa/Ga + Pb/Gb)
Gsb = (65 + 35) / (65/2.45 + 35/3.25)
Gsb = (65 + 35) / 37.299843
Gsb = 100 / 37.299843
Gsb = 2.68
How to Cancel prescription
The purpose of the international residential code is to
Answer:
The International Building Code (IBC) is a model code that provides minimum requirements to safeguard the public health, safety and general welfare of the occupants of new and existing buildings and structures.
Explanation:
can help me with this circuit question?
the last word is (cutoff)
Answer:
The answer is "[tex]25\times 10^{-9}[/tex]".
Explanation:
Using formula:
[tex]f_c=\frac{1}{2\pi RC}\\\\w_c= 4 \frac{krad}{sec}\\\\w_c=2\pi fc\\\\R=w\\\\c=\frac{1}{w_c\ R}\\\\[/tex]
[tex]=\frac{1}{4 \times 10^3 \times 10\times 10^3}\\\\=\frac{1}{40 \times 10^6 }\\\\=0.025 \times 10^{-6 }\\\\=25\times 10^{-9}[/tex]
explain all the characteristics of computer
A turbine of a fossil fuel burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1790 V, what current does it deliver (in A)
Answer:
The generator delivers current of 500.11 A
Explanation:
Given the data in the question;
mechanical energy delivered to the generator = 1500 hp
efficiency η = 80.0 %
terminal potential difference of the generator = 1790 V
we know that;
1 hp = 746 W
so
the mechanical energy delivered to the generator will be
Generator Input = ( 1500 × 746 )W = 1119000 W
So the generator output will be;
Generator Output = Generator Input × η
we substitute
Generator Output = 1119000 W × 80.0 %
Generator Output = 1119000 W × 0.8
Generator Output = 895200 W
So the Current will be;
[tex]I[/tex] = Generator Output / terminal potential difference of the generator
we substitute
[tex]I[/tex] = 895200 W / 1790 V
[tex]I[/tex] = 500.11 A
Therefore, The generator delivers current of 500.11 A
Can some people answer these questions so i can get to know the age group i an making my target market for DT-GSCE thankyou if you do my deadline is tomorrow :D
Answer:
I think you might have forgotten to post the problems
Can you use isentropic efficiency for a non-adiabatic compressor?
Can you use isothermal efficiency for an adiabatic compressor?
Technician A says white smoke in the exhaust of a diesel engine can be the result of a cylinder misfire in a warm engine. Technician B says blue smoke in the exhaust of a diesel engine can be caused by scored cylinder walls. Who is correct?
Answer:
Both
Explanation:
Because of water, fuel does not burn completely. This brings about water fumes that are white in color and looks like white smoke. If engine is cold and water is heating, it leads to steam formation like water vapor. The white times are because of not firing properly in the heated engine. Technician A is right.
Blue fine is caused by this scoring. It is also caused by dirty oil. Technician b is right too
A cylindrical block of wood 1 m in diameter and 1 m long has a specific weight of 7500 N/m^3. Will it float in water with its axis vertical?
Answer:
The block will float with its axis vertical.
Explanation:
For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.
So, weight of cylindrical block, W = specific weight × volume
specific weight = 7500 N/m³
volume = πd²h/4 where d = diameter of block and h = height of block
volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³
W = 7500 N/m³ × 0.7854 m³ = 5890.5 N
Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³
Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced = 9800 N/m³ × 0.7854 m³ = 7696.92 N
Since W' = 7696.92 N > W = 5890.5 N, the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.
Resistance depends on which three properties of a wire?
Color and texture are not directly related to a wire’s resistance.
1. color, thickness, texture
2. thickness, length, temperature
3. length, texture, temperature
4. temperature, color, texture
Answer:
2
Explanation:
From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.
A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.
Answer:
The answer is "22.501,-22.899"
Explanation:
Just as in the previous problems find the angle the velocity makes with the x-axis and radius of curvature.
[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]
For the radius of curvature, we can use the expression from the last two problems, but first express the position and derivatives as y(x).
[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]
The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)
[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]
Safety Issues for Operators of Oil and Gas Exploiting Equipment when working off rigs
Answer:
Safety First, Safety Always. Safety stands out as a core value for the oil and natural gas industry, embedded in every process and decision for operations. The oil and natural gas industry and the federal government are working together to continuously improve the safety of offshore operations. ...
A ceramic specimen with an elastic modulus of 300 GPa is under a tensile stress of 800 MPa. Will it fracture if its most severe flaw is an internal crack of 0.30 mm long with a tip radius of curvature in the amount of 0.0015 mm? Please justify your conclusion. (Hint: Compare the largest stress in the specimen around the crack to the theoretical strength which is roughly E/10).
Answer:
16Gpa < 30 Gpa
there would be no fracture
Explanation:
fracture can occur if the maximum strength at the top of the biggest flaw is more than the theoretical fracture
to get the theoretical strength =
e/10 = 300/10
= 30 Gpa
we get the magnitude at the buggest flaw
= 2σ√a/ρt
σ = 800
ρτ = 0.0015
a= 0.3/2
[tex]=2*800\sqrt{\frac{\frac{0.3}{2} }{0.0015} }[/tex]
= [tex]=2*800*\sqrt{100} \\=2*800*10\\=16000MPa[/tex]
= 16Gpa < 30 Gpa
the fracture is not going to happen given that the maximum strenght is smaller than the theoretical fracture strength.
If you deposit $ 1000 per month into an investment account that pays interest at a rate of 9% per year compounded quarterly.how much will be in your account at the end of 5 years ?assume no interpèriod compounding
Answer:
5,465.4165939453
Explanation:
formula
A=P(1+r/n)^n(t)
p=1000
r=0.09
n=4
t=5
Which of the following is NOT part of a car's drive train?
A axle
B rotor
C differential
D transmission
Answer:
B. rotor
Explanation:
The correct answer Is rotor because the others are part of a cars drivetrain
The drive train system exists as a critical element of a vehicle and the transmission exists as an integral part of the drive train. B rotors NOT part of a car's drive train.
Which is not part of the drive train?A drive train exists not really a single part of your car – it's a set of drive train components that interact with the engine to drive the wheels and different regions of the vehicle to thrust it into motion. These components often contain the transmission, differential, driveshaft, axles, CV joints, and wheels.
Therefore, the correct answer is option B rotors.
To learn more about the part of a car's drive train
https://brainly.com/question/14362401
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A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium.If the left and bottom edges of the plate are fixed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate.
Answer:
[tex]X_t=2.17391304*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]e_t=0.0026[/tex]
[tex]e_r=0.0035[/tex]
Explanation:
From the question we are told that:
Dimension [tex]12*12[/tex]
Thickness [tex]l_t=5mm=5*10^-3[/tex]
Normal tensile force on top side [tex]F_t= 15kN[/tex]
Normal tensile force on right side [tex]F_r= 20kN[/tex]
Elastic modulus, [tex]E=115Gpap=>115*10^9[/tex]
Generally the equation for Normal Strain X is mathematically given by
[tex]X=\frac{Force}{Area*E}[/tex]
Therefore
For Top
[tex]X_t=\frac{Force_t}{Area*E}[/tex]
Where
[tex]Area=L*B*T[/tex]
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}[/tex]
[tex]X_t=2.17391304*10^{-4}[/tex]
For Right side[tex]X_r=\frac{Force_r}{Area*E}[/tex]
Where
Area=L*B*T
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
Generally the equation for elongation is mathematically given by
[tex]e=strain *12[/tex]
For top
[tex]e_t=2.17391304*10^{-4}*12[/tex]
[tex]e_t=0.0026[/tex]
For Right
[tex]e_r=2.89855072*10^{-4} *12[/tex]
[tex]e_r=0.0035[/tex]
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Engineers design products or processes to meet desired needs. Your desired need or goal (hopefully) is to graduate with your Bachelor of Science degree in engineering. But what is the process you need to apply to be successful in achieving this goal?
The number of pulses per second from IGBTs is referred to as
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm and 5.0 cm, find the volume flow rate and the average velocity in each pipe section.
Answer:
volumetric flow rate = [tex]0.0251 m^3/s[/tex]
Velocity in pipe section 1 = [tex]6.513m/s[/tex]
velocity in pipe section 2 = 12.79 m/s
Explanation:
We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.
The density of water is = 997 kg/m³
density = mass/ volume
since we are given the mass, therefore, the volume will be mass/density
25/997 = [tex]0.0251 m^3/s[/tex]
volumetric flow rate = [tex]0.0251 m^3/s[/tex]
Average velocity calculations:
Pipe section A:
cross-sectional area =
[tex]\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2[/tex]
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
[tex]velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s[/tex]
Pipe section B:
cross-sectional area =
[tex]\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2[/tex]
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
[tex]velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s[/tex]
A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show that buoy does not float with its axis vertical. (II). What minimum pull should be applied to a chain attached to the center of the base to keep the buoy vertical?
Answer:
[tex]GM<0[/tex]
So the bouy does not float with its axis vertical
Explanation:
From the question we are told that:
Diameter [tex]d=2m[/tex]
Length [tex]l=2.5m[/tex]
Weight [tex]W=22kN[/tex]
Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]
Generally the equation for weight of cylinder is mathematically given by
Weight of cylinder = buoyancy Force
[tex]W=(pwg)Vd[/tex]
Where
[tex]V_d=\pi/4(d)^2y[/tex]
Therefore
[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]
Therefore
Center of Bouyance B
[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]
Center of Gravity
[tex]G=\frac{I.B}{2}=2.6m[/tex]
Generally the equation for\BM is mathematically given by
[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]
Therefore
[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]
Therefore
[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]
Therefore
[tex]GM<0[/tex]
So the bouy does not float with its axis vertical
Why does the ceramic made from Thorium and Oxygen have the chemical ratio of 2 oxygen atoms to every thorium atom (ThO2)