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STALIN [3.7K]
3 years ago
9

Sand particles carried by wind can grind against the surface of rocks. This is called _____. abrasion hydrolysis oxidation depos

ition
Physics
2 answers:
Ulleksa [173]3 years ago
3 0
Abrasion

hope this helps
Lemur [1.5K]3 years ago
3 0

Answer:

abrasion

Explanation:

Abrasion is the grinding away of rock by friction and impact during transportation. Rocks are broken down into smaller particles by the mechanism of weathering. There after erosion causes the transportation of these rock particles.The rocks particles began to mechanically scrape themselves due to friction . This process of rocks scraping themselves to causing wearing on the surface is known as Abrasion. The rock particles might be transported by wind, glacier, waves and erosion.

Due to abrasion the rock particles might further disintegrate into more smaller particles.    

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

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}

\theta=53.13^o

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Answer:

5694000 min

Explanation:

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