All categories

4 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.

You might be interested in

The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/

velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

To get the true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯, P⃗ =AC¯¯¯¯¯¯¯¯

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector. To get the coordinates of these two vectors, use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

8 0

3 years ago

If i am changing velocity, i must also have _______ and a net _________

Alinara [238K]

if i am changing velocity, i must also have acceleration and a net force

<h2>Newton's first law of motion</h2>

Newton's first law of motion states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

According to Newton's first law of motion, without a force acting on an object, its velocity does not change. The net force acts on an object to change its velocity and cause acceleration.

Read more about velocity:

brainly.com/question/4931057

6 0

3 years ago

4. The Mariana trench is in the Pacific Ocean and has a depth of approximately 11,000 m. The density of seawater is approximatel

Alex73 [517]

Explanation:

It is known that relation between pressure and density is as follows.

P = $\rho gh$

where, P = pressure

$\rho$ = density

g = acceleration due to gravity

h = height

Putting the given values into the above formula as follows.

P = $\rho gh$

= $1025 \times 9.8 \times 11000$

= 110495000 Pa

Now, relation between pressure and force is as follows.

P = $\frac{F}{A}$

or, F = PA

F = $110495000 \times \pi \times (0.1)^{2}$

= $3.47 \times 10^{6} N$

Thus, we can conclude that a force of $3.47 \times 10^{6} N$ can be experienced at such depth.

3 0

3 years ago

In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$