What is the effect on a field of tall grass when a wave caused by a cool summer breeze passes through the field?

1 answer:

mihalych1998 [28]1 year ago

7 0

Whenever a wave is brought on by a cool summer breeze across the field. The grass tilts in the wind's direction before reverting to its original posture. Option A is correct.

<h3>What is the wind blowing?</h3>

Air pressure variations within our atmosphere result in the wind, which is the movement of air.

Areas with a low pressure are approached by air in high pressure. The air flows more quickly when there is a bigger pressure differential. Most rapid winds.

The wind is moving air and is caused by differences in air pressure within our atmosphere.

Air under high pressure moves toward areas of low pressure. The greater the difference in pressure, the faster the air flows. The Fastest Winds.

Hence, option A is correct.

To learn more about the wind blowing refer;

brainly.com/question/14212137

#SPJ1

You might be interested in

Which of the following statements is true?

klemol [59]

Answer: The more massive an object, the faster it will accelerate.

Explanation:

4 0

2 years ago

6. Which country in the world is frequently effected by earth quakes?

prohojiy [21]

Answer:

For which country do we locate the most earthquakes? Japan. The whole country is in a very active seismic area, and they have the densest seismic network in the world, so they are able to record many earthquakes.

plz mark as brainlist

7 0

2 years ago

An object is swung in a horizontal circle on a length of string

Genrish500 [490]

Answer: either 3.20s or 1.18s

Explanation:

4 0

3 years ago

An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel

Karo-lina-s [1.5K]

I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door $d$ )...

The door forms a right triangles that satisfies

$\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}$

We also have

$\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd$

so if you happen to know the height of the door, you can solve for $b$ and $a$ .

$d$ is fixed, so

$a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}$

We can solve for the angular velocity $\dfrac{\mathrm d\theta}{\mathrm dt}$ :

$\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}$