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tangare [24]
3 years ago
13

If they opened up the International Space Station to tourism, would you go? Why? answer in 2 sentences

Engineering
2 answers:
Arlecino [84]3 years ago
6 0
Personally, I would go to the space station. The space station has extreme different levels of technology and abilities, plus who doesn’t want to go to space.
nydimaria [60]3 years ago
4 0

Answer:

I’d go to Uranus.

Explanation:

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A master stud pattern is laid out somewhat<br> like a?
Svetradugi [14.3K]

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

3 0
2 years ago
The Accenture team is involved in helping a client in the transformation journey using Cloud computing. How is myNav beneficial
professor190 [17]

Answer:

It helps ensure the security of client data.

Explanation:

Cloud computing is one of the security features in networking which is being adopted by big corporations and organizations inorder to ensure smooth running of the organization.

Also, due to the sensitivity of data which some organizations deals in, they tried everything possible to protect themselves and their clients' information through use of cloud computing.<em> Data are stored in the clouds, and requires special administrative rights for it to be accessed by some of the staffs working in any given organization.</em>

8 0
3 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
3 years ago
A _____ satellite system employs many satellites that are spaced so that, from any point on the Earth at any time, at least one
Wittaler [7]

Answer:

d. low earth orbit (LEO)

Explanation:

This type of satellites form a constellation deployed as a series of “necklaces” in such a way that at any time, at least one satellite is visible by a receiver antenna, compensating the movement due to the earth rotation.

Opposite to that, a geostationary satellite is at an altitude that makes it  like a fixed point over the Earth´s equator, rotating synchronously with the Earth, so it is always visible in a given area.

3 0
3 years ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
Radda [10]

Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

4 0
4 years ago
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