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earnstyle [38]
4 years ago
14

Which of the following is considered a calm area close to the equator resulting in little to no wind?

Physics
1 answer:
solong [7]4 years ago
5 0
<h2>Right answer: Doldrums</h2>

These are also called zones of equatorial calm and it is due a climatic phenomenon that is placed near the Earth equator, attributed to the soft winds, that are called calm winds as well; accompanied by systems of abundant rains and heat.

In this area periods of great calm occur when the winds virtually disappear completely, trapping the sailing ships for long periods (days or weeks). This is why the term <em>doldrum</em> became popular as a colloquial expression in the eighteenth century, to refer to "<em>the caprice of the wind that slows down the navigation to sail". </em>


The zone is located in the place where two trade winds meet, this means the trade winds of the northern hemisphere <u>converge</u> with those of the southern hemisphere, that is why this region is related to the <u>intertropical convergence zone</u>.


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The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Evgen [1.6K]

Question Continuation

Derive an expression for x in terms of m, M, and D. b) If the net force is zero a distance ⅔D from the planet, what is the ratio R of the mass of the planet to the mass of the moon, M/m?

Answer:

a. x = (D√M/m)/(√M/m + 1)

b. The ratio R of the mass of the planet to the mass of the moon=4:1

Explanation:

Given

m = Mass of moon

M = Mass of the planet

D = Distance between the centre of the planet and the moon

Net force = 0

Let Y be a point at distance x from the planet

Let mo = mass at point Y

a.

Derive an expression for x in terms of m, M and D.

Formula for Gravitational Force is

F = Gm1m2/r²

Y = D - x

Force on rest mass due to mass M (FM) =Force applied on rest mass due to m (Fm)

FM = G * mo * M/x²

Fm = G * mo * m/Y²

Fm = G * mo * m/(D - x)²

FM = Fm = 0 ------ from the question

So,

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- Cross Multiply

M * (D - x)² = m * x²

M/m = x²/(D - x)² ---_ Find square roots of both sides

√(M/m) = x/(D - x) ----- Multiply both sides by (D - x)

(D - x)√(M/m) = x

D√(M/m) - x√(M/m) = x

D√(M/m) = x√(M/m) + x

D√(M/m) = x(√(M/m) + 1) ------- Divide both sides by √M/m + 1

x = (D√M/m)/(√M/m + 1)

b. Here x = ⅔D

FM = G * mo * M/x²

Fm = G * mo * m/(D - x)²

FM = Fm

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- (Substitute ⅔D for x)

M/(⅔D)² = m/(D - ⅔D)²

M/(4D/9) = m/(⅓D)²

9M/4D = m/(D/9)

9M/4D = 9m/D ---- Divide both side by 9/D

M/4 = m

M = 4m

M/m = 4

M:m = 4:1

So, the ratio R of the mass of the planet to the mass of the moon=4:1

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3 years ago
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I believe it is kinetic energy
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