An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal.
The water cycle is all about storing water and moving water on, in, and above the Earth. Although the atmosphere may not be a great storehouse of water, it is the superhighway used to move water around the globe. Evaporation and transpiration change liquid water into vapor, which ascends into the atmosphere due to rising air currents. Cooler temperatures aloft allow the vapor to condense into clouds and strong winds move the clouds around the world until the water falls as precipitation to replenish the earthbound parts of the water cycle. About 90 percent of water in the atmosphere is produced by evaporation from water bodies, while the other 10 percent comes from transpiration from plants.
There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.
Answer:
<span>GPE=81000J or 81kJ</span>
Explanation
Potential Energy = mgh = 20 x 9.8 x ?
<span>To find H use one of the equation of motion </span>
<span>= [(90)^2 - 0 ] / 2(9.8) </span>
<span>Potential Energy = mgh = 20 x 9.8 x 8100 /2(9.8) = 81000 J</span>
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
