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Bingel [31]
2 years ago
11

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

Physics
1 answer:
mihalych1998 [28]2 years 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

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Answer:

Essentials. A battery is a device that stores chemical energy and converts it to electrical energy. The chemical reactions in a battery involve the flow of electrons from one material (electrode) to another, through an external circuit. The flow of electrons provides an electric current that can be used to do work.

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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an
agasfer [191]

Answer:

t=\sqrt{h/g}

Explanation:

We use the kinematics equation to solve this question:

y(t)=y_{o}+v_{o}t+1/2*a*t^{2}

v_{o}=0    because the ball is dropped

a=-g         the acceleration is the gravity, negative because it points downwards

y_{o}=h     initial height

y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

t=\sqrt{h/g}

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Two groups of students were tested to compare their speed working math problems. Each group was given the same problems. One gro
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