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3 years ago

How do sea surface temperatures affect evaporation rate?

1 answer:

8 0

Answer: The temperature doesn't affect the evaporation rate, but affects on how much of water a parcel of air can contain when saturated which is known by the absolute humidity. Hurricanes are usually happening when the temperature of the sea water west of the Cape Verde islands is over 27 degrees Celsius. If ahead of the path of a hurricane, the sea water temperature drops then it will be less moisture in the air and perhaps the hurricane will fade out. But it is not as simple. How strong a tropical storm is is relative to the difference of temperture between ground level and the top of the troposphere. The greater the difference, the faster the air will rise and the deeper the pressure will be, forcing surrounding air to rush in, thus forming a hurricane force wind. Then there is the fact that the wet adiabatic lapse rate is about half that of dry air. It means that rising moist air cools down slower and therefore rises higher. Hence water is the true fuel of bad weather. But it can't be isolated from the fact that the difference of temperature must be great too. What we often forget is that the tropopause (the border to the stratosphere) is much higher over the equator and therefore, much colder than e.g. the poles.

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An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k

ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

$g=\frac{GM}{R^{2}}$

$g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}$

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

$v^{2}=u^{2}-2gh$

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0

2 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight W, with magnitude W = 100 N, pointing directly downward

• two tension forces T₁ and T₂ with equal magnitude T₁ = T₂ = T = 1000 N, both making an angle of θ with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ F = T₁ sin(θ) + T₂ sin(180° - θ) - W = 0

We have sin(180° - θ) = sin(θ) for all θ, so the above reduces to

2T sin(θ) = W

2 (1000 N) sin(θ) = 100 N

sin(θ) = 0.05

θ ≈ 2.87°

If y is the vertical distance between the stoplight and the ground, then

tan(θ) = (15 m - y) / (100 m)

Solve for y :

tan(2.87°) = (15 m - y) / (100 m)

y = 15 m - (100 m) tan(2.87°)

y ≈ 9.99 m

3 0

2 years ago

A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa

Fantom [35]

Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute = $W_B$ =10 g

Given mass of solvent= $W_A$ =90 g

$k_f=1.86^{\circ}C/m$

Freezing point of solution =-3.33 $^{\circ}$ C

Freezing point of solvent = $0^{\circ}$ C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33= $3.33^{\circ}$

$\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}$

$M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}$

$M_B=62.06 g/mol$

Hence, molar mass of unknown liquid is 62.06g/mol.

6 0

2 years ago

Read 2 more answers

A force of 16 N is applied to a 2 kg mass over 4 m. The surface is horizontal and frictionless. If the object is initially at re

KATRIN_1 [288]

C) Calculate the final speed.

6 0

2 years ago

The product of m1m2 in the formula for gravitational force is always

Nataly [62]

Answer:

The correct answer is the second option

Explanation:

The formula in question (for gravitational force) is provided below

F = G.m₁.m₂/r²

Where

F = force of gravity

G = gravity constant (6.674 × 10⁻¹¹ m.kg⁻¹.s⁻²)

m₁ and m₂ = are the masses of the objects

r² = distance between the centers of the two objects

From the above, it can be deduced that m₁ and m₂ are masses and hence can only be positive. Positives multiplied together only gives positive, thus the second option is correct.

8 0

3 years ago