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masha68 [24]
3 years ago
6

Which type of erosion and deposition is most common in coastal areas around the Gulf of Mexico

Physics
2 answers:
Alja [10]3 years ago
7 0
<span> most common in coastal areas around the gulf of mexico is mostly by river

</span>
Law Incorporation [45]3 years ago
7 0

water erosion and deposition is most common in coastal areas around the Gulf of Mexico. this erosion and deposition is mainly caused by rain, river and storms. it is the combination of two process named as detachment and travelling of particles.

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Newton’s cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes
Marysya12 [62]

Answer: A

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2 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
2 years ago
Read 2 more answers
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Weight is the force exerted on an object by gravity. So, weight of any object on the moon is 1/6 that on Earth.
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Please tell the correct answer​
alukav5142 [94]

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The answer is C she didn't mark C so its your best best of trying.Good Luck!

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What is the major regulator of vocal intensity?
Burka [1]
<span>The major regulator of vocal intensity is the lung pressure.

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