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3 years ago

When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?

Physics

1 answer:

Anarel [89]3 years ago

3 0

Answer: If we have equilibrium, the magnitude must be zero.

Explanation:

If the charges are in equilibrium, this means that the total charge is equal to zero.

And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero

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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:

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3 years ago

What is the kinetic energy of a 0.5 kg puppy that is running 1.5m/s

bija089 [108]

Answer: Well the answer is KE = 5.625E-7 i just don't know the units for it...

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3 years ago

Which characteristic gives the most information about what kind of element an atom is ?

Butoxors [25]

Answer:

The atomic number

Explanation:

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3 years ago

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Why are high tides found simultaneously on opposite sides of earth?

dexar [7]

Explanation:

Why are high tides found simultaneously on opposite sides of Earth? The tidal bulges occur on both sides of Earth that are aligned with the tide-generating body. The ocean water experiencing high tide rotates around Earth on a 12-hour cycle.

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3 years ago

Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t

Tamiku [17]

Answer:

Approximately $9.62$ .

Explanation:

$y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0]$ .

$y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)]$ .

Notice that sine waves $y_1$ and $y_2$ share the same frequency and wavelength. The only distinction between these two waves is the $(-0.250)$ in $y_2\!$ .

Therefore, the sum $(y_1 + y_2)$ would still be a sine wave. The amplitude of $(y_1 + y_2)\!$ could be found without using calculus.

Consider the sum-of-angle identity for sine:

$\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)$ .

Compare the expression $\sin(a + b)$ to $y_2$ . Let $a = (4.35\, x - 1270)$ and $b = (-0.250)$ . Apply the sum-of-angle identity of sine to rewrite $y_2\!$ .

$\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Therefore, the sum $(y_1 + y_2)$ would become:

$\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Consider: would it be possible to find $m$ and $c$ that satisfy the following hypothetical equation?

$\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Simplify this hypothetical equation:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}$ .

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}$ .

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real $x$ and $t$ , the following should be satisfied:

$\displaystyle 1 + \cos(-0.250) = m\, \cos(c)$ , and

$\displaystyle \sin(-0.250) = m\, \sin(c)$ .

Consider the Pythagorean identity. For any real number $a$ :

${\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2$ .

Make use of the Pythagorean identity to solve this system of equations for $m$ . Square both sides of both equations:

$\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2$ .

$\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2$ .

Take the sum of these two equations.

Left-hand side:

$\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}$ .

Right-hand side:

$\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}$ .

Therefore:

$m^2 = 2 + 2\, \cos(-0.250)$ .

$m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98$ .

Substitute $m = \sqrt{2 + 2\, \cos(-0.250)}$ back to the system to find $c$ . However, notice that the exact value of $c\!$ isn't required for finding the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ .

(Side note: one possible value of $c$ is $\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125$ radians.)

As long as $\! c$ is a real number, the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ would be equal to the absolute value of $(4.85\, m)$ .

Therefore, the amplitude of $(y_1 + y_2)$ would be:

$\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}$ .

8 0

3 years ago