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horsena [70]
3 years ago
5

Summarize ocean acidification in one sentence.

Physics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

The ocean absorbs a significant portion of carbon dioxide (CO2) emissions from human activities, equivalent to about one-third of the total emissions for the past 200 years from fossil fuel combustion, cement production and land-use change (Sabine et al., 2004). Uptake of CO2 by the ocean benefits society by moderating the rate of climate change but also causes unprecedented changes to ocean chemistry, decreasing the pH of the water and leading to a suite of chemical changes collectively known as ocean acidification. Like climate change, ocean acidification is a growing global problem that will intensify with continued CO2 emissions and has the potential to change marine ecosystems and affect benefits to society.

The average pH of ocean surface waters has decreased by about 0.1 unit—from about 8.2 to 8.1—since the beginning of the industrial revolution, with model projections showing an additional 0.2-0.3 drop by the end of the century, even under optimistic scenarios (Caldeira and Wickett, 2005).1 Perhaps more important is that the rate of this change exceeds any known change in ocean chemistry for at least 800,000 years (Ridgewell and Zeebe, 2005). The major changes in ocean chemistry caused by increasing atmospheric CO2 are well understood and can be precisely calculated, despite some uncertainty resulting from biological feedback processes. However, the direct biological effects of ocean acidification are less certain

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1 “Acidification” does not mean that the ocean has a pH below neutrality. The average pH of the ocean is still basic (8.1), but because the pH is decreasing, it is described as undergoing acidification.

Page 2

Suggested Citation:"Summary." National Research Council. 2010. Ocean Acidification: A National Strategy to Meet the Challenges of a Changing Ocean. Washington, DC: The National Academies Press. doi: 10.17226/12904. ×

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and will vary among organisms, with some coping well and others not at all. The long-term consequences of ocean acidification for marine biota are unknown, but changes in many ecosystems and the services they provide to society appear likely based on current understanding (Raven et al., 2005).

In response to these concerns, Congress requested that the National Research Council conduct a study on ocean acidification in the Magnuson-Stevens Fishery Conservation and Management Reauthorization Act of 2006. The Committee on the Development of an Integrated Science Strategy for Ocean Acidification Monitoring, Research, and Impacts Assessment is charged with reviewing the current state of knowledge and identifying key gaps in information to help federal agencies develop a program to improve understanding and address the consequences of ocean acidification (see Box S.1 for full statement of task). Shortly after the study was underway, Congress passed another law—the Federal Ocean Acidification Research and Monitoring (FOARAM) Act of 2009—which calls for, among other things, the establishment of a federal ocean acidification program; this report is directed to the ongoing strategic planning process for such a program.

Although ocean acidification research is in its infancy, there is already growing evidence of changes in ocean chemistry and ensuing biological impacts. Time-series measurements and other field data have documented the decrease in ocean pH and other related changes in seawater chemistry (Dore et al., 2009). The absorption of anthropogenic CO2 by the oceans increases the concentration of hydrogen ions in seawater (quanti-

Explanation:

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Rapid gas exchange can be accomplished more easily in air than in water because A. the O2 content of water is higher than that o
bagirrra123 [75]

Answer:

C. water is more dense and viscous

Explanation:

Rapid gas exchange can be accomplished more easily in air than in water because water is more dense and viscous.

Gases have the greatest ease of diffusion of their respective particles, as occurs in air, since their molecules have higher speeds and have more distance from each other than liquids.

The molecular diffusion rate in liquids is much less than in gases. The molecules of a liquid are very close (liquids are more dense and viscous) to each other compared to those of a gas, then the gas molecules hits with the molecules of the liquid with more frequency and this causes that the gas moves slower than in other gas (for example in air).

3 0
3 years ago
An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.
amm1812

Answer:

a) T ’= 0.999 s ,  b)  t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

        w = √g / L

The angular velocity, frequency and period are related

        w = 2π f = 2π / T

        2π / T = √ g / L

        T = 2π √ L / g

        L = T² g / 4π²

        L = 1² 9.8 / 4π²

        L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

       ΔL = α L ΔT

       ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

       ΔL = 142.8 10⁻⁶ m

       Lf - L = -142. 8 10⁻⁶

       Lf = 142.8 10⁻⁶ + 0.248

       Lf = 0.2479 m

Let's calculate new period

      T ’= 2π √ L / g

      T ’= 2π √ (0.2479 / 9.8)

      T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

       t = 3600 0.999

       t = 3596.4 s

Therefore the clock is delayed almost 4 s

6 0
4 years ago
A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of
Simora [160]

Answer:

Approximately 1.5\; \rm kg. (Assuming that this table is level, and that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}.

Explanation:

Convert the dimensions of this book to standard units:

\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.30\; \rm m.

\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.25\; \rm m.

Calculate the surface area of this book:

0.30\; \rm m  \times 0.25\; \rm m = 0.075\; \rm m^{2}.

Pressure is the ratio between normal force and the area over which this force is applied.

\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}.

Equivalently:

(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area}).

In this question, \text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}.

It was found that (\text{contact Area}) = 0.075\; \rm m^{2} (assuming that the entire side of this book is in contact with the table.

Hence:

\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}.

If that the table is level, this normal force would be equal to the weight of this book:

\text{weight} = 15\; \rm N.

Assuming that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}. The mass of this book would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}.

4 0
3 years ago
A (n) ______ is a material that transfers heat well
Vanyuwa [196]
A (n) conductor is a material that transfers heat well

Hope that helped
4 0
4 years ago
Read 2 more answers
A 780 g , 54-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short b
evablogger [386]

Answer:

The angular velocity is <em>35.5 rad/s.</em>

Explanation:

Step 1: Find the moment of Inertia for the rod

The moment of inertia, <em>I</em>, of a rod:

<em>I=\frac{1}{12} ML^{2} .......... (1)</em>

where M = mass of rod (0.78 Kg); L = Length of rod (0.54 m).

I=\frac{1}{12} (0.78)(0.54)^{2} =0.019\frac{Kg}{m^{2} }

Step 2: Calculate the angular acceleration from Rotational kinetic notation

<em>F.r = I.α ........... (2)</em>

<em>where F is the force acting upon the rod; r is the half length of the rod; I is the moment of Inertia and; α is the angular acceleration.</em>

<em>∴ (1000 N)(0.27 m) = 0.019α</em>

<em>α = 270 Nm / 0.019 Kgm²</em>

<em>α = 14210.5 rad/s</em>

Step 3: We find the angular velocity by using the equation below:

<em>ωf = ωi + αt ......... (3)</em>

<em>where </em>

<em>ωf is the angular velocity after the blow</em>

<em>ωi is the angular velocity before the blow = 0</em>

<em>t is the time taken for the blow to occur = 2.5 ms</em>

<em>ωf = 0 + (14210.5 rad/s)(2.5 ms) = 35.5 rad/s.</em>

<em />

6 0
3 years ago
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