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3 years ago

What is the source of the energy released in fission and fusion reactions?

Physics

2 answers:

Alex Ar [27]3 years ago

7 0

Strong nuclear forces

Explanation:

The formation and breakage of the strong nuclear forces in the nucleus of an atom accounts for the energy released during fission and fusion reactions.

Nuclear reactions occurs when changes are made to the nucleus of an atom.

Strong nuclear forces maintains the stability in the nucleus of an atom.
Strong nuclear forces are the strongest of the fundamental forces.
They hold subatomic particles together.
When they are broken during a nuclear fission, they release a great deal of energy.
When they form during a fusion they also release energy.

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blagie [28]3 years ago

3 0

Answer:

mass converted to energy

Explanation:

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

Inga [223]

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:

7 0

3 years ago

Which process generates energy in the Sun?1) nuclear fusion2) nuclear fission3) chain reaction4) transmutation

Ber [7]

1) nuclear fusion

During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. During this process, radiant energy is released.

5 0

1 year ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

$a_c = 3.2 g$

So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

$\alpha = \frac{a_t}{r}$

where r = 10.9 m is the length of the centrifuge. Substituting,

$\alpha = \frac{29.6}{10.9}=2.7 rad/s^2$

Learn more about centripetal and angular acceleration here:

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8 0

3 years ago

List 3 to 5 ways to reduce friction

pychu [463]

<span>Make the surfaces smoother. Rough surfaces produce more friction and smooth surfaces reduce friction
Lubrication is another way to make a surface smoother
Make the object more streamlined
Reduce the forces acting on the surfaces
<span>Reduce the contact between the surfaces.</span></span>

8 0

3 years ago

Read 2 more answers

Which quantity below is a derived quantity?

Reil [10]

By definition, the speed of an object is given by:

$v = \frac{dr}{dt}$