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omeli [17]
3 years ago
15

Select the correct answer

Engineering
1 answer:
kifflom [539]3 years ago
5 0

Answer: C. Exchanging ideas with the group will help build his knowledge and reputation

Explanation:

Professional networking refers to the building of professional relationships. This helps in establishing connections that are mutually-beneficial with the people that are in ones profession.

Based on the scenario given, the reason for Vijay to join a networking group is that exchanging ideas with the group will help build his knowledge and reputation.

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A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M
Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of 300\times 100 mm^2

Force(F)=8 MN

Therefore stress due to this Force(\sigma)

\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}

\sigma =266.66 \times 10^{6} Pa

\sigma =266.66 MPa

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

8 0
3 years ago
Analyze the example of this band saw wheel and axle. The diameter of the wheel is 14 inches. The diameter of the axle that drive
Kazeer [188]

The answer for the ideal mechanical advantage and actual mechanical advantage for the different scenarios are;

A) Ideal Mechanical Advantage = 18.67

B) Actual Mechanical Advantage = 4.1067

We are given;

Input distance; The diameter of the wheel; d_w = 14 inches

Output distance; The diameter of the axle that drives the wheel; d_a = 3/4 inches

The force needed to cut a one-inch-thick softwood board; F = 1.75 pounds

The efficiency of the band saw; η = 22% = 0.22

A) Formula for Mechanical advantage is;

M.A = Force output/Force input = (Input distance)/(Output distance)

Thus;

Ideal mechanical advantage = 14/(3/4)

Ideal mechanical advantage = 18.67

B) Now, we are given that efficiency of the band saw is η = 22% = 0.22.

Thus using the mechanical advantage formula above;

Actual mechanical advantage = 0.22 × Expected output

Actual mechanical advantage = 0.22 × 18.67

Actual mechanical advantage ≈ 4.1067

Read more about Mechanical Advantage at; brainly.com/question/18345299

5 0
2 years ago
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Answer:

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5 0
3 years ago
Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:
natta225 [31]

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

7 0
3 years ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

8 0
3 years ago
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