Large bodies of water<span> such as oceans, seas, and large lakes </span>affect<span> the </span>climate<span> of an area. </span>Water<span> heats and cools more slowly than land. Thus, in the summer, the </span>coastal<span> regions </span>will<span> stay cooler and in winter warmer. A more moderate </span>climate<span> with a smaller temperature range </span>is<span> created.</span>
Answer:
It would be lower than because, if the boiling point of that element is 77 Kelvin degrees then if it isn’t at boiling point it would automatically be cooler than that. Even if it’s at its original state. The normal temperature of Liquid Nitrogen is really cold -320.8 degrees.
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Answer:
Correct answer: t = 2.86 seconds
Explanation:
We first use this formula
V² - V₀² = 2 a d
where V is the final velocity (speed), V₀ the initial velocity (speed),
a the acceleration and d the distance.
We will calculate the acceleration from this formula
a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10
a = 0.525 m/s²
then we use this formula
V = V₀ + a t => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds
t = 2.86 seconds
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Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
