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andreev551 [17]
3 years ago
13

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and

had final mass of 0.490 kg. Determine dry bulk density, porosity and water content. 10 pts
Engineering
2 answers:
belka [17]3 years ago
8 0

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of  sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = \frac{wet mass - dry mass }{wet mass} = \frac{0.525 - 0.49}{0.525} = 0.07 or 7%

water content =  \frac{wet mass - dry mass}{wet mass} = 7%

amm18123 years ago
8 0

Answer:

Given diameter 4cm=0.04m,lenght=0.12m

Iniatial mass m1=0.525kg final mass m2=0.490kg

dry bulk density =?

Porosity=?

Water content=?

Water content

Mass of water=0.525-0.490=0.035kg

m=Mw/Ms=0.935)0.525=0.0667 Pr 6.67%

Dry bulk density

Recall Density= mass/volume

Volume =πr^2h=π*(0.02)^2*0.12=7.54*10^-3m^3

Dry density=0.490/(7.54*10^-3)=64.97kg

Porosity

n=e/1+e

Were void ratio e=0.0667*2.6=0.1734

Porosity=0.1734/1+0.1734=0.15

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A fluid flowing along a flat plate will stick to it at the point of contact

Explanation:

and this is known as the no-slip condition. ... This is the precise reason why shear stress in a fluid can also be interpreted as the flux of momentum.

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What statement exemplifies the role of archeology in architecture?
horrorfan [7]
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Look at the home page of the Internet Society (www.internetsociety.org) and read about one of the designers of the original ARPA
krek1111 [17]

Answer:

<u>ARPANET is the direct precedent for the Internet, a network that became operational in October 1969 after several years of planning. </u>

Its promoter was DARPA (Defense Advanced Research Projects Agency), a US government agency, dependent on the Department of Defense of that country, which still exists.

Originally, it connected research centers and academic centers to facilitate the exchange of information between them in order to promote research. Yes, being an undertaking of the Department of Defense, it is understood that weapons research also entered into this exchange of information.

It is also explained, without being without foundation, that the design of ARPANET was carried out thinking that it could withstand a nuclear attack by the USSR and, hence, probably the great resistance that the network of networks has shown in the face of major disasters and attacks.

It was the first network in which a packet communication protocol was put into use that did not require central computers, but rather was - as the current Internet is - totally decentralized.

Explanation:

<em><u> Below I present as a summary some of the most relevant aspects exposed on the requested website about the origin and authors of ARPANET:</u></em>

<em><u></u></em>

1. Licklider from MIT in August 1962 thinking about the concept of a "Galactic Network". He envisioned a set of globally interconnected computers through which everyone could quickly access data and programs from anywhere. In spirit, the concept was very much like today's Internet. He became the first head of the computer research program at DARPA, and from October 1962. While at DARPA he convinced his successors at DARPA, Ivan Sutherland, Bob Taylor and MIT researcher Lawrence G. Roberts, of the importance of this network concept.

2.Leonard Kleinrock of MIT published the first article on packet-switching theory in July 1961 and the first book on the subject in 1964. Kleinrock convinced Roberts of the theoretical feasibility of communications using packets rather than circuits, That was an important step on the road to computer networking. The other key step was to get the computers to talk together. To explore this, in 1965, working with Thomas Merrill, Roberts connected the TX-2 computer in Mass. To the Q-32 in California with a low-speed phone line creating the first wide-area (albeit small) computer network built . The result of this experiment was the understanding that timeshare computers could work well together, running programs and retrieving data as needed on the remote machine, but that the circuitry switching system of the phone was totally unsuitable for the job. Kleinrock's conviction of the need to change packages was confirmed.

3.In late 1966 Roberts went to DARPA to develop the concept of a computer network and quickly developed his plan for "ARPANET", and published it in 1967. At the conference where he presented the document, there was also a document on a concept of UK packet network by Donald Davies and Roger Scantlebury of NPL. Scantlebury told Roberts about NPL's work, as well as that of Paul Baran and others at RAND. The RAND group had written a document on packet switched networks for secure voice in the military in 1964. It happened that work at MIT (1961-1967), in RAND (1962-1965) and in NPL (1964-1967) all they proceeded in parallel without any of the investigators knowing about the other work. The word "packet" was adopted from the work in NPL and the proposed line speed to be used in the ARPANET design was updated from 2.4 kbps to 50 kbps.

6 0
3 years ago
How many people made machines
olganol [36]

Answer:

The total number of people whom have made machines is not a recorded figure? Need to be more specific :/

Explanation:

Sorry not very helpful, your question is REALLY broad

3 0
3 years ago
Read 2 more answers
Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Interne
dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0
3 years ago
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