Ask question

Ask question

All categories

Dahasolnce [82]

3 years ago

7

Precipitation tends to be ___ on the _ side of a mountain because water vapor _____ as it rises there.

1 answer:

Lera25 [3.4K]3 years ago

6 0

Answer:

Higher, Windward side, Condenses

Explanation:

The Windward side refers to that side of a mountain that faces the direction from which the wind is blowing. In this direction, the moisture containing hot air blowing from a distant place moves upward and strikes the mountain at a greater height, where the air mass is thin and the temperature is relatively cold. As the temperature and pressure decrease with altitude, the hot uprising air cools and gradually condenses. This results in the occurrence of high precipitation over this region i.e. the windward side of the mountain.

Therefore, the precipitation is always higher on the windward side of a mountain as the hot air undergoes condensation at greater height as it rises upward.

You might be interested in

Can someone help me with science:

Oliga [24]

Answer:

trur

Explanation:

gvidttcyhjddgdhjfjdjsjs

6 0

3 years ago

Read 2 more answers

What is the answer for this question (4x³)²

Delvig [45]

16x to the power of 6

6 0

3 years ago

Read 2 more answers

What type of general wave requires a medium for the wave to travel through

Gemiola [76]

Answer:

Mechanical

Explanation:

Electromagnetic waves are waves that have no medium to travel whereas mechanical waves need a medium for its transmission.

7 0

2 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

Why are there multiple versions of the scientific method?

gavmur [86]

Answer:

it's because some versions have more steps and others have less

4 0

2 years ago