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Softa [21]
3 years ago
9

Familiarize yourself with the map showing the DSDP Leg 3 drilling locations and the position of the mid-ocean ridge (Figure 1 to

the right). Using the latitude and longitude coordinates on the sides of the map, and this online map for reference, in what ocean were these samples collected?
Physics
1 answer:
Inga [223]3 years ago
7 0

Answer:

For more than 40 years, results from scientific ocean drilling have contributed to global understanding of Earth’s biological, chemical, geological, and physical processes and feedback mechanisms. The majority of these internationally recognized results have been derived from scientific ocean drilling conducted through three programs—the Deep Sea Drilling Project (DSDP; 1968-1983), the Ocean Drilling Program (ODP; 1984-2003), and the Integrated Ocean Drilling Program (IODP; 2003-2013)—that can be traced back to the first scientific ocean drilling venture, Project Mohole, in 1961. Figure 1.1 illustrates the distribution of drilling and sampling sites for each of the programs, and Appendix A presents tables of DSDP, ODP, and IODP legs and expeditions. Although each program has benefited from broad, international partnerships and research support, the United States has taken a leading role in providing financial continuity and administrative coordination over the decades that these programs have existed. Currently, the United States and Japan are the lead international partners of IODP, while a consortium of 16 European countries and Canada participates in IODP under the auspices of the European Consortium for Ocean Research Drilling (ECORD). Other countries (including China, Korea, Australia, New Zealand, and India) are also involved.

As IODP draws to a close in 2013, a new process for defining the scope of the next phase of scientific ocean drilling has begun. Illuminating Earth’s Past, Present, and Future: The International Ocean Discovery Program Science Plan for 2013-20231 (hereafter referred to as “the science plan”), which is focused on defining the scientific research goals of the next 10-year phase of scientific ocean drilling, was completed in June 2011 (IODP-MI, 2011). The science plan was based on a large, multidisciplinary international drilling community meeting held in September 2009.2 A draft of the plan was released in June 2010 to allow for additional comments from the broader geoscience community prior to its finalization. As part of the planning process for future scientific ocean drilling, the National Science Foundation (NSF) requested that the National Research Council (NRC) appoint an ad hoc committee (Appendix B) to review the scientific accomplishments of U.S.-supported scientific ocean drilling (DSDP, ODP, and IODP) and assess the science plan’s potential for stimulating future transformative scientific discoveries (see Box 1.1 for Statement of Task). According to NSF, “Transformative research involves ideas, discoveries, or tools that radically change our understanding of an important existing scientific or engineering concept or educational practice or leads to the creation of a new paradigm or field of science, engineering, or education. Such research challenges current understanding or provides pathways to new frontiers.”3 This report is the product of the committee deliberations on that review and assessment.

HISTORY OF U.S.-SUPPORTED SCIENTIFIC OCEAN DRILLING, 1968-2011

The first scientific ocean drilling, Project Mohole, was conceived by U.S. scientists in 1957. It culminated in drilling 183 m beneath the seafloor using the CUSS 1 drillship in 1961. During DSDP, Scripps Institution of Oceanography was responsible for drilling operations with the drillship Glomar Challenger. The Joint Oceanographic Institutions for Deep Earth Sampling (JOIDES), which initially consisted of four U.S. universities and research institutions, provided scientific advice. Among its numerous achievements, DSDP

Explanation:

You might be interested in
A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo
DIA [1.3K]

Answer:

74^{\circ} C

Explanation:

We are given that

Mass of glass,m=300 g

T_1=23^{\circ}

Volume,V=236cm^3

Mass of water=density\times volume=1\times 236=236 g

Density of water=1g/cm^3

Temperature of hot water,T=87^{\circ}

Specific heat of glass,C_g=0.2cal/g^{\circ}C

Specific heat of water,C_w=1 cal/g^{\circ}C

Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)

Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)

Q_{glass}+Q_{water}=0

300\times 0.2(T_f-23)+236\times 1(T_f-87)

60T_f-1380+236T_f-20532=0

296T_f=20532+1380=21912

T_f=\frac{21912}{296}=74^{\circ} C

5 0
2 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
Would the forces acting on the sky diver be balanced or unbalanced? Explain your answer.
docker41 [41]

Answer:

your mom is balanced!!!

4 0
2 years ago
From the following statements about mechanical waves, identify those that are true for transverse mechanical waves only, those t
Dominik [7]

'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.

'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.

'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.

<u>Explanation:</u>

Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.

As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.

So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.

And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.

5 0
2 years ago
The total negative charge on the electrons in 1mol of helium (atomic number 2, molar mass 4) is ________?
DaniilM [7]

Answer:

1.92\times 10^5 C

Explanation:

We are given that

Atomic number=2

We have to find the total negative charge on the electrons in one mole of Helium.

We know that atomic number=Proton number

Proton number=Number of electrons=2

Number of electrons in Helium=2

1 mole of Helium=6.02\times 10^{23} atoms

We know that q=ne

Where n =Number of fundamental units

e=Charge on electron

1 e=1.6\times 10^{-19}C

Using the formula

q=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C

Total negative charge in 1 mole=3.2\times 10^{-19}\times 6.02\times 10^{23}=1.92\times 10^5C

Hence, the total negative charge on the electrons in 1 mole of Helium=1.92\times 10^5 C

4 0
3 years ago
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