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2 years ago

Find the True statement

Engineering

2 answers:

dmitriy555 [2]2 years ago

6 0

Answer:

Option A is correct ( Stress relaxation is time- and temperature-dependent).

Explanation:

Luda [366]2 years ago

6 0

Answer:

Option A is correct.

The only completely true statement in all of the available options is that "Stress relaxation is time- and temperature-dependent"

Explanation:

Stress relaxation is the constant decrease of stress with time in a material which is exposed to a constant strain at a constant temperature.

It is similar in concept to vaccination; in that, the material is being prepared to be able to undergo a particular amount of stress at a specific temperature in order to make the stress levels in the material lower than it usually should be.

It is evident that stress relaxation, (the extent), in a material completely depends on the amount of time that the constant strain is applied and the temperature at which this whole process occurs.

Polymers always have a melting temperature.

While it is true that almost all materials can melt. For Polymers, it is not necessarily true that the melting takes place at a constant temperature. In very complex Polymers, melting occurs over an appreciable range of temperature, not sharply as in some simple polymers.

Hence, polymers don't always have a melting temperature, some polymers melt over a range of temperature. So, this statement isn't completely true.

Polyethylene becomes weaker if H is replaced with other radicals inside branches.

This also is a statement that isn't always true. Replacing the H in some of the branches of polyethylene with radicals such as the carboxylic group, the hydroxyl group etc., have been shown to seriously strengthen the polyethylene. Polymers formed by replacing the H with ion-Like radicals are known as ionomers.

These groups have a polarity, hence, they are drawn together by their charges, come together in microdomains, toughening and strengthening the polyethylene without taking its ability to be cast to permanent shapes away.

This statement isn't always true too.

Crazing indicates cracks inside the polymer.

This is also false. Crazing doesn't indicate cracks in materials. What it does is that Crazing precedes the cracks.

Crazing is a phenomenon that entails the formation of microvoids in a material in response to excessive tensile stress being applied to the material. The microvoids (which forms in a plane normal to the tensile stress that caused them) often leads to cracks or fracture later.

So, Crazing doesn't indicate cracks, rather, it precedes them in most cases.

Hope this Helps!!!

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name the process by which mild steel can be converted into high carbon steel and explain it briefly ?

Fudgin [204]

Answer:

please give me brainlist and follow

Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

6 0

2 years ago

Jack has been concerned about the rapidly changing green regulations in his state and his ability as a mechanical engineer to ke

uysha [10]

Answer:

Option A, B and D

Explanation:

Jack can easily convince boss if he focus around two major aspects of the company

a) Revenue enhancement - Jack must outline the benefits of his research that can be used to improvise customer offerings and hence can be further used to devise more energy-efficient options to customer

b) Reduction in mistakes - Issues such as poor implementation can be avoided with better approach and understanding.

Hence, option A, B and D are correct

3 0

2 years ago

Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum

ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F?

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

Make the assumption Base continuous flow velocity (BFFS)= 65 mph. 

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = Vp /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

Vp = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * fHV * fP) = 2668.5

fHV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (ET -1)) = 0.754

ET = 4.26 ~ 3.5

For 4% upgrade and 10% trucks with ET = 3.5, length of the grade is Greater than 1.0 miles

6 0

3 years ago

Read 2 more answers

Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured

aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

$CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]$

If $\bar{x}$ is 190, we can find the value of $z\sigma_{\bar{x}}$ :

$\bar{x} - z\sigma_{\bar{x}} = 188.29$

$190 - z\sigma_{\bar{x}} = 188.29$

$z\sigma_{\bar{x}} = 1.71$

Now we need to find the value of $\sigma_{\bar{x}}$ :

$\sigma_{\bar{x}} = s / \sqrt{n}$

$\sigma_{\bar{x}} = 5/ \sqrt{25}$

$\sigma_{\bar{x}} = 1$

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

$CI = 1 - 2*0.0436 = 0.9128 = 91.28\%$

The nearest CI in the options is 90%, so the correct option is B.

4 0

3 years ago

A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature = - 35 degree C

Maximum expansion can be determined as

$\Delta L = L \alpha (T_{max} - T_{min})$

where , \alpha is expansion coefficient $= 12\times 10^{-6}$ degree C

SO, $\Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))$

$\Delta L = 0.45 m = 450 mm$

number of minimum expansion joints is calculated as

$n = \frac{450}{20} = 22.5$

b) length of each bridge

$Length = \frac{500}{22.5} = 22.22 m$

8 0

2 years ago