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garik1379 [7]
3 years ago
13

The upward and downward movement of air in the atmosphere creates pressure systems. Arrange the steps in the correct order to de

scribe how a high-pressure system forms and leads to pleasant weather. The skies clear, making way for pleasant weather. Cool, dense air begins to sink downward. Pressure differences cause wind to blow outward. A high-pressure system forms.
Physics
2 answers:
notka56 [123]3 years ago
6 0

Answer:

Cool, dense air begins to sink downward.

A high-pressure system forms.

Pressure differences cause wind to blow outward.

The skies clear, making way for pleasant weather.

Explanation:

OleMash [197]3 years ago
4 0

Answer:

  1. Cool, dense air begins to sink downward.
  2. A high-pressure system forms.
  3. Pressure differences cause wind to blow outward.
  4. The skies clear, making way for pleasant weather.

Explanation:

The air near the land warms up and rises. Cool dense air sinks being heavier. This creates a pressure difference. A high pressure system forms. This difference leads to blowing of wind from High pressure to low pressure area. Eventually sky gets clear and we experience pleasant weather. Hence, the correct order to describe this is:

  1. Cool, dense air begins to sink downward.
  2. A high-pressure system forms.
  3. Pressure differences cause wind to blow outward.
  4. The skies clear, making way for pleasant weather.
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Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

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Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

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