Winds blowing across the ocean surface push water away. Water then rises up from beneath the surface to replace the water that was pushed away. This process is known as “upwelling.”
Upwelling occurs in the open ocean and along coastlines. The reverse process, called “downwelling,” also occurs when wind causes surface water to build up along a coastline and the surface water eventually sinks toward the bottom.
Water that rises to the surface as a result of upwelling is typically colder and is rich in nutrients. These nutrients “fertilize” surface waters, meaning that these surface waters often have high biological productivity. Therefore, good fishing grounds typically are found where upwelling is common.
A.) The higher the altitude, the colder the climate will be
B.) Areas near the equator have warmer climates than areas for form the equator.
D.) Winds that blow inland from oceans or large lakes contain a lot of water vapor that will cause precipitation.
C.) Monsoons.
Answer:
These are the two statements with scientific facts that explain the described phenomenon
<span>
Gravitation between two objects increases when the distance between them decreases.</span>
When the mass of an object increases, its gravitational pull also increases.
Justification:
Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>
That law is represented by the equation:
F = G × m₁ × m₂ / d²
The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.
The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.
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Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m
