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timofeeve [1]
3 years ago
13

List five things the welding symbol will tell the welder about the weld that is to be made.​

Engineering
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

depth

rod to use

travel speed

type of weld

tthe side to be welded on

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate h
shusha [124]

Mahamad Siddiqui sent false emails and letters of recommendation on behalf of individuals without their permission to nominate himself for the Waterman Award at the National Science Foundation. His earlier emails were offered where he had solicited letters were offered as evidence. Siddiqui claimed that content of earlier emails was hearsay. Do the earlier emails come in is given below

Explanation:

1.Mohamed Siddiqui appeals his convictions for fraud and false statements to a federal agency, and obstruction in connection with a federal investigation.   Siddiqui challenges the district court's admission into evidence of e-mail and foreign depositions.

2.On February 18, 1997, Jodi Saltzman, a special agent with the NSF interviewed Siddiqui at Siddiqui's office at the University of South Alabama.   During the interview, Siddiqui signed a statement admitting that he had nominated himself for the Waterman Award, but that he had permission from Yamada and von Gunten to submit forms on their behalf.   Siddiqui also acknowledged in the statement that Westrick had recommended Siddiqui for a different award, the PECASE Award, but that Siddiqui had changed the wording of the letter to apply to the Waterman Award.   Siddiqui was indicted on April 29, 1997.

3.Siddiqui opposed the taking of the depositions on the grounds that the witnesses' personal presence at trial was necessary, and that Indian travel restrictions for its citizens residing abroad prevented him from traveling to Japan and Switzerland.   Specifically, Siddiqui asserted that because of religious persecution in India his travel to Japan or Switzerland related to the criminal action would put his family members still living in India at risk.   The magistrate judge ruled that the government had carried its burden of showing that Yamada and von Gunten would be unavailable to appear at trial, and instructed that Siddiqui's fear of obtaining a travel visa from India because of the threat of persecution of family members should not preclude the taking of the foreign depositions.

4.Yamada's deposition was taken in Japan on March 6, 1998.   At government expense, Siddiqui's counsel attended the deposition and cross-examined the witness, but was not in telephonic contact with Siddiqui during the deposition.   Yamada testified that on February 1, 1997, she received an e-mail stating that if she received a phone call from the NSF to “please tell good words about me.”   Yamada testified that she knew the e-mail was from Siddiqui because the name on the e-mail had Siddiqui's sender address, and it ended with the name “Mo” which Siddiqui had previously told her was his nickname, and which he had used in previous e-mail.

5.Yamada later admitted to Saltzman that she had not given Siddiqui permission to sign, but had made the earlier representation because she thought Siddiqui would go to jail.

6.During cross-examination of Yamada at the deposition, Siddiqui's counsel introduced an e-mail from Yamada to Siddiqui.   This e-mail contained the same e-mail address for Siddiqui as the e-mail received by Yamada and von Gunten apparently from Siddiqui.

7.Von Gunten's video deposition was taken in Switzerland.   At government expense, Siddiqui's counsel attended the deposition and cross-examined von Gunten.   During the deposition, Siddiqui was in communication with his counsel by telephone.   Von Gunten testified at the deposition that he had not submitted a letter of recommendation in favor of Siddiqui for the Waterman Award, and that he had not given Siddiqui permission to submit such a letter in his name.

8 0
4 years ago
a stem and leaf display describes two-digit integers between 20 and 80. for one one of the classes displayed, the row appears as
allochka39001 [22]

Answer:

  52, 50, 54, 54, 56

Explanation:

The "stem" in this scenario is the tens digit of the number. Each "leaf" is the ones digit of a distinct number with the given tens digit.

  5 | 20446 represents the numbers 52, 50, 54, 54, 56

8 0
3 years ago
Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
What scale model proves the initial concept?
Tju [1.3M]

Answer: A prototype

Explanation:

5 0
3 years ago
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