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KiRa [710]
3 years ago
5

A cylindrical specimen of this alloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm. 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 so, calculate the load. If not, explain why.
Engineering
1 answer:
Margarita [4]3 years ago
5 0

Answer:

Condition 2 is true.

\epsilon_{test}>\epsilon_{yield}

we can not calculate the load with given information.

According to above values, Condition 2 is satisfied so we can not find the load with given information because material deformation lies in plastic region.

Explanation:

In order to check whether we can find the load or not we have to check the following conditions:

Condition 1:

\epsilon_{test}

If this condition is true then we can calculate the load.

Condition 2:

\epsilon_{test}>\epsilon_{yield}

If this condition is true then we can not calculate the load with given information.

Calculating \epsilon_{test}:

\epsilon_{test}=\frac{Elongation}{original\ length}

\epsilon_{test}=\frac{7.6\ mm}{250\ mm}\\ \epsilon_{test}=0.0304

Calculating \epsilon_{yield}:

\epsilon_{yield}=\frac{Yield\ Stress}{Elastic\ modulus}\\ \epsilon_{yield}=\frac{280\ Mpa}{105*10^3\ MPa} \\\epsilon_{yield}=2.67*10^{-3}=0.00267

Hence:

Condition 2 is true.

\epsilon_{test}>\epsilon_{yield}

According to above values, Condition 2 is satisfied so we can not find the load with given information because material deformation lies in plastic region.

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Answer:

  d. all of these

Explanation:

Electrostatic discharge will generally produce excess voltage in a local area that results in excessive current and excessive heat. It will blast a crater in an MOS device, or melt bond wires, or cause damage of other sorts. In short, MOS devices are subject to damage from "all of these."

6 0
3 years ago
A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo
Kazeer [188]

Answer:

volume  = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'\times H

where A' = 61% of A

              = 0.61\times 2667 = 1626.87 sq ft

volume  =  1626.87 \times (\frac{14}{12} ft)

               =1898.015 ft^3

in\ m^3 = \frac{ 1898.015}{35.315} =   53.7457 m^3

in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon

weight = \rho Vg

       = 1000\times 53.74\times 9.8

             =526652 N

In\ lbf =  \frac{526652}{4.448} = 118396.08 lbf

7 0
3 years ago
What is an air mass?​
kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0
3 years ago
Read 2 more answers
4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD
babunello [35]

Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

5 0
3 years ago
Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
2 years ago
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