An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 2.01cm and the young's modulus of aluminum is 69 GPa

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Answer 1

The question is incomplete. The complete question is --

An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :

1. the interference value, i

2. the radial pressure at the interference of the collar and the rod

3. the maximum effective stress in the collar

4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break

Solution:

Given :

Inner radius of the collar = 1 cm

So, inner diameter, [tex]$d_1$[/tex] = 2 cm

Outer radius of the collar = 12 cm

So, outer diameter, [tex]$d_2$[/tex] = 4 cm

The aluminium rod diameter, d = 1.01 cm

Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.

1. So, the interference , i =0

2. The radial pressure is also 0.

3. There will be no stress developed. So the maximum effective stress is 0

4. The collar will not break


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Discuss the aging that occurs in asphalt cement during mixing with aggregates and in service. How can the different types of aging of asphalt cement be simulated in the laboratory?

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Answer:

It can be simulated with RTFO and PAV method

Explanation:

The aging which occurs for the binders is always tested using rolling thin film oven and also pressure aging vessels. The rolling thin film oven is a method used to calculate the short term aging that occurs in the asphalt cement when it mixes with the aggregates as well as in service. While the long term process of aging in asphalt cement is carried out by pressure aging vessel.

The short term process of aging can be simulated with the help of RTFO and the long term is simulated by the PAV method.

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Explanation:

The purpose of this program is to recognize the Youth's vital role in nation- building, promote consciousness among youth and develop their physical, moral, spiritual, intellectual and social well-being. It shall inculcate in the youth patriotism, nationalism, and advance their involvement in public and civic affairs.

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The regulations require various records to be kept for a minimum period of 3 years. Such records must be kept at the place where the work to which they relate is carried on or, if this is not practicable, at the responsible persons place of business. Such records must include The date and time of the occurrence.

A rigid, insulated vessel is divided into two equal-volume compartments connected by a valve. Initially, one compartment con tains 1 m3 of water at 20°C, x = 50%, and the other is evacuated. The valve is opened and the water is allowed to fill the entire volume. For the water, determine the final temperature, in °C, and the amount of entropy produced, in kJ/K.

Answers

Solution :

Given

Volume, [tex]$V_1 = 1 \ m^3$[/tex]

Temperature, [tex]$T_1=20 \ ^\circ C[/tex]

[tex]$x_1=0.5$[/tex]

From the saturated water table, corresponding to [tex]$T_1=20 \ ^\circ C[/tex], we get the saturated liquid, vapor specific and the entropy.

[tex]$v_f=1.0010 \ m^3/kg$[/tex]

[tex]$v_g=57.791 \ m^3/kg$[/tex]

[tex]$s_f=0.2966 \ kJ/kg-K$[/tex]

[tex]$s_g=8.6672 \ kJ/kg-K$[/tex]

Now calculating the initial specific volume

[tex]$v_1=v_f+x_1 \cdot(v_g-v_f)$[/tex]

    [tex]$=1.0018+05 \cdot(57.791-1.0018)$[/tex]

    [tex]$= 29.8973 \ m^3/kg$[/tex]

Calculating the initial specific entropy:

[tex]$s_1=s_f+x+1 \cdot (s_g-s_f)$[/tex]

    [tex]$=0.2966+0.5 \cdot (8.6673 - 0.2966)$[/tex]

    [tex]$= 4.48 \ kJ/kg-K$[/tex]

So final volume of the vessel is two times bigger as the initial volume

[tex]$V_2=2 .V_1$[/tex]

    [tex]$= 2 \times 29.8973 = 59.8 \ m^3/kg$[/tex]

If we interpolate the values from tables between [tex]$v_g=57.791 \ m^3/kg$[/tex] and [tex]$v_g=61.293 \ m^3/kg$[/tex], we can get final temperature and specific entropy corresponding to value of [tex]$v_2$[/tex] :

Final temperature, [tex]$T_2= 19.6 ^\circ C$[/tex]

and [tex]$s_2 = 8.68 \ kJ/kg-K$[/tex]

Calculating change in entropy

[tex]$\Delta s = s_2-s_1$[/tex]

     [tex]$=8.68-4.48 = 4.2 \ kJ/kg-K$[/tex]

What is the process of a Diesel engine uses to convert fuel to mechanical energy

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Answer:

A diesel engine is a type of heat engine that uses the internal combustion process to convert the energy stored in the chemical bonds of the fuel into useful mechanical energy. ... First, the fuel reacts chemically (burns) and releases energy in the form of heat.

A soil sample, taken from a borrow pit has a specific gravity of soil solids of 2.66. The sample was taken to a materials laboratory and tested. The results of a standard Proctor test are tabulated below.

Weight of Soil (lb) Moisture Content (%)
3.20 12.8
3.78 13.9
4.40 15.0
4.10 15.7
3.70 16.6
3.30 18.1

The maximum dry density in lb/ft3 is most nearly:_______

Answers

Answer:

115 Ib/ft^3

Explanation:

To determine the maximum dry density in Ib/ft3 we have to calculate :

Bulk unit weight ( yb ) ; W / v

Dry unit weight ( yd. ) :  yb / ( 1 + w )

For every set of data given

assuming  v = 1/30 ft^3

calculating for the 3 data set ( maximum dry density )

weight of soil (W) = 4.40

moisture content (%) (w) = 15.0 = 0.15

Bulk unit weight (yb) = 4.40 / (1/30)  = 132 Ib/ft^3

Dry unit weight ( yd. ) = 132 / ( 1 + 0.15 ) = 114.702 Ib/ft^3

therefore after calculations the maximum dry density in Ib/ft^3 ≈ 115 Ib/ft^3

A typical printed page of text contains 50 lines of 80 characters each. Imagine that a certain printer can print 6 pages per minute and that the time to write a character to the printer’s output register is so short it can be ignored. Does it make sense to run this printer using interrupt‐driven I/O if each printer requires an interrupt that takes 50 µsec all‐in to service?

Answers

Solution :

Given

The number of line that can be written by printer per page = 50 lines

The number of characters that can be written by printer per page = 80 characters.

Number of pages that can be written per minute = 6 pages

The speed for writing characters to the printer can be found by

= 50 x 80 x 6

= 400 characters per sec

In one second the printer has the writing speed of 400 character.

Now each of the character uses = 50 µsec of the CPU time for interrupt.

So in each second, the interrupt overhead = 20 msec

Now using the interrupt driven input/output,  980 msec time can be available for other work.

Thus, the interrupt overhead charges only 2 percent of the CPU, that will hardly affect the program to run.

10. Which one of the following items would you expect to find on a floor framing plan?
ОООО
A. Sizes of joists
O B. Height of objects
C. Location of plumbing features
D. Arrangement of heating features

Answers

Answer:

c

Explanation:

the pluming usually is on the bottom floor plan

The answer is c hope it helps

sup 100 point for you

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Answer:

thanks! :D

Have a good day!

Answer:

Yay, :)

Thank you!!!

When do you need to apply for program completion and review?

Answers

Answer:

1/2 semesters before completion

Explanation:

Program completion and review are both a necessary part and parcel of a successful completion of any program. The recommended period for applying for program completion and review is one to two semesters before the program is to be completed. This is mandatory, so that there is different time span for review and any other issues that are needed to be carried out. And then afterwards, finally the certificates are then issued to the qualified person.

How are glasses a form of technology and how do they involve engineering aspects?

Please help!!!!

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I’m not sure what type of answer you would like but this is the one that I’m going to give you they are a part of architecture aspect because they or a process of making from every screw to the glass from finding the perfect mold just imagine how you would make them

will mark brainliest if correct
When a tractor is driving on a road, it must have a SMV sign prominently displayed.

True
False

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Answer: true

Explanation:

Explain the major differences between designing buildings for earthquake resistance and for wind resistance.

Answers

Answer:

Earthquake's put vibration and seismic load on a building while wind puts a continuous force throughout the whole structure.

Explanation:

The most important difference between the design of a building that's resistent to earthquakes and resistant to high speed winds is that earthquake's put vibration and seismic load on a building while wind puts a continuous force throughout the whole structure.

Earthquake resistent building are equipped with a base that is isolated from the buildings with several layers of different material which absorb the vibration of the seismic waves and prevent them from reaching to the building.

Wind resistent buildings must have strong foundations and a sturdy structure/material that can withstand high pressure that the wind applies during events such as hurricanes.

I hope this answer helps.  

​What should be a concern as a weldment becomes larger as more parts are added?

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Which company produces comprehensive patch management software?
Quest
Motorola
Apple
Sony

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If I’m not mistaken it is Quest. I could be completely wrong or I could be right. I really don’t know but I guess you’ll have to go with it if you choose my answer. But Quest seems to be correct.

The cross section of a heat exchanger consists of three circular pipes inside a larger pipe. The internal diameter of the three smaller pipes is 2.5 cm, and the pipe wall thickness is 3 mm. The inside diameter of the larger pipe is 8 cm. If the velocity of the fluid in the region between the smaller pipes and larger pipe is 10 m/s, what is the discharge in m^3/s?

Answers

Answer:

0.0432 m^3/s

Explanation:

Internal diameter of smaller pipes = 2.5 cm = 0.025 m

pipa wall thickness = 3 mm = 0.003 m

internal diameter of larger pipes = 8 cm = 0.8 m

velocity of region between smaller and larger pipe = 10 m/s

Calculate discharge in m^3/s

First we calculate the area of the smaller pipe

A = [tex]\pi Dt[/tex] = [tex]\pi ( 0.025 ) ( 0.003 )[/tex]  = 0.00023571 m^2

next we calculate area of fluid between the smaller pipes and larger pipe

A = [tex][\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})[/tex]

   = [tex][ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )][/tex]

   = [ 0.00502857 - 0.00070713 ]

   = 0.00432144 m^2

hence the discharge in m^3/s

Q = AV

   = 0.00432144 * 10

   = 0.0432 m^3/s

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the t's are in minutes and the x's are in thousands of vehicles per hour. The travel times on the routes are known to be in user equilibrium. If an observation for route 1 finds that the gaps between 30% of the vehicles are less than 6 seconds. Estimate the volume and average travel times for the two routes

Answers

Solution :

Given

[tex]$t_1=2+x_1$[/tex]

[tex]$t_2=1+x_2$[/tex]

Now,

[tex]$P(h<5)=1-P(h \geq5)$[/tex]

[tex]$0.4=1-P(h \geq5)$[/tex]

[tex]$0.6=P(h \geq5)$[/tex]

[tex]$0.6= e^{\frac{-x_1 5}{3600}}$[/tex]

Therefore,   [tex]$x_1=368 \ veh/h$[/tex]

                        [tex]$=\frac{368}{1000} = 0.368$[/tex]

Given,   [tex]$t_1=2+x_1$[/tex]

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, [tex]$t_2=t_1$[/tex]

∴  [tex]$t_2$[/tex] = 2.368 min

[tex]$t_2=1+x_2$[/tex]

[tex]$2.368=1+x_2$[/tex]

[tex]$x_2 = 1.368$[/tex]

[tex]$x_2 = 1.368 \times 1000$[/tex]

    = 1368 veh/h

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Answer:

i'll help you but there is no question to answer??

2) The switch in the circuit below has been closed a long time. At t=0, it is opened.
Find the inductor current for il(t) for t> 0.

Answers

Answer:

  il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

  il(t) = e^(-t/.01)

  il(t) = e^(-100t) . . . amperes

Lower headlight beams must be used when approaching within __________ of an oncoming vehicle or when following within __________ of the rear of another vehicle.

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Answer:

500 feet, 300 feet

Explanation:

How are you today First to awnser is brainlyest.

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Answer:

Happy

Explanation:

Discoloration on walls, work surfaces, ceilings, walls, and pipes may indicate a leak that is causing you to waste raw materials.

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Answer:

True :)

Explanation:

If this is a true or false question.

The discoloration is when the color of a particular thing comes off or changes, fading from its original color. In other words, when the original color of anything changes or wanes off, then that is called discoloration. The statement is true.

What is Discoloration?

This chemical shift occurs for a number of reasons, including its interaction with other elements nearby or the unfavorable effects of other elements it comes into touch with.

Water leaks may create cases of discoloration on walls, surfaces, even ceilings, and pipes. As a result, the surface progressively changes and becomes moist.

Discoloration puts the surrounding region in danger in addition to altering the part(s) or area's look and durability.

Since a change in that region or component is required, the raw material(s) used are wasted. The claim that discoloration results in the waste of raw materials are thus accurate.

Learn more about discoloration here:

brainly.com/question/6378613

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Oxygen enters an insulated 12-cm-diameter pipe with a velocity of 70 m/s. At the pipe entrance, the oxygen is at 240 kPa and 20°C, and at the exit it is at 200 kPa and 18°C. Calculate the rate at which entropy is generated in the pipe.

Answers

Answer:

S = 0.10253 kW/k

Explanation:

Given data:

Velocity of oxygen ( V ) = 70 m/s  

Diameter of pipe = 12-cm = 0.12 m

At entrance

pressure of oxygen ( p1 ) = 240 kPa

Temperature of oxygen ( T1 ) = 20°c  = 293 k

At exit

pressure of oxygen ( p2 ) = 200 kPa

temperature of oxygen ( T2 ) = 18°c = 291 k

First calculate specific volume of oxygen at inlet

V1 = [tex]\frac{RT1}{P1}[/tex]  -------- ( 1 )

  R = 0.2598 KJ/kgk ( property of oxygen )

  T1 = 293 k

  P1 = 240 kpa

substitute values into equation 1

V1 = 0.3172 m^3/kg

next we calculate the mass flow rate of Oxygen

m = [tex]\frac{A1V}{v1}[/tex] ----- ( 2 )

A1 ( area of pipe ) = 0.0113 m^2 ( calculated )

V = 70 m/s

V1 = 0.3172 m^3/kg

substitute value into equation 2

m ( mass flow rate of oxygen ) = 2.4936 kg/s

Finally calculate the rate at which entropy is generated in the pipe

S = [tex]m( C_{p} In\frac{T2}{T1} - RIn\frac{P2}{P1} )[/tex]   --------- ( 3 )

[tex]C_{p}[/tex] = 0.918 kj/kgk  ( property of oxygen )

T2 = 291 k

T1 = 293 k

P2 = 200 kPa

P1 = 240 kPa

substitute values into equation 3 above

S = 0.10253 kW/k

what do you expect the future trends of an operating system in terms of (a) cost (b) size (c) multitasking (d) portability (e) simplicity​

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Answer:

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Explanation:

The set of line styles and line thickness used on drawings used on drawings are referred to as the ____.

Answers

Answer:

I think it's referred as the outline

I think it’s the outline but I’m not too sure

Igneous rocks are formed due to the

a cooling of magma
b weathering of rocks
c changes in pressure
d deposition of sediment

Answers

A. Cooling of magma
Would be the answer

Answer:

a. cooling of magma

Explanation:

they form when magma cools

Find the minimum sum of products expression using Quine-McCluskey method of the function. F(A,B,C,D)= Σ m(1,5,7,8,9,13,15)+ Σ d(4,6,11)

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Answer:

Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.

Explanation:

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For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of alloy that is liquid.

Answers

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

             wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= [tex]\frac{R}{R +S } * 100[/tex]

=  [tex]\frac{53-45}{58-45} * 100[/tex]  = 61.5%

To determine the wt.% of the alloy that is liquid

= [tex]\frac{S}{S+R} * 100\\[/tex]

= [tex]\frac{58-53}{58-45} *100[/tex] = 38.5%

attached below is a free hand sketch as well

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