(Write the answer in fair test) You have to investigate whether surface area affects how fast some water evaporates

5. A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 6400 m away?

EastWind [94]

Answer:

8.89 secs

Explanation:

720m = 1 sec

6400m = ?

6400/720 = 8.89 secs

8 0

3 years ago

Summarize ocean acidification in one sentence.

Snowcat [4.5K]

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:

3 0

3 years ago

Help Plsss

sattari [20]

On the Newtonian theory of gravity, gravitation affects anything with mass. Assuming that none of the answer choices is the only thing that exists in the universe, all of the answer choices are subject to the law of universal gravitation (hence “universal”).

Satellites, water, frogs, and stars all have mass as they are all composed of matter. Thus, all four answer choices should be circled.

3 0

4 years ago

Which learning style describes children who use the body effectively like a surgeon or a dancer?

creativ13 [48]

Answer:

A. Bodily Kinesthetic

Explanation:

3 0

3 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at t = 4.7 s will be:

v = v0 + a · t (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = 4.2 m/s

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = 9.9 m

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x = v · t

x = 4.2 m/s · 11.7 s = 49.1 m

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t² (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0

3 years ago