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Vesnalui [34]
2 years ago
5

Help me!

Physics
2 answers:
Marysya12 [62]2 years ago
5 0
It is a compound yep
ratelena [41]2 years ago
4 0

Answer:D.Compound

Explanation: internet

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Water that flows across the surface is called a;

Runoff

That's when rain has saturated the ground to the point it cant hold anymore and it runs over the surface.

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during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which f
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Can we use momentum to see how fast the earth is going?
Kisachek [45]

Yes, if we know the Earth's mass

Explanation:

The momentum of an object is a vector quantity given by the equation

p=mv

where

m is the mass of the object

v is its velocity

In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

v=\frac{p}{m}

Therefore, if we know the momentum of the Earth (p) and we know its mass as well (m), we can solve the formula to find the Earth's velocity.

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3 years ago
In the diagram, which is demonstrated by arrow 1? A. liquid water evaporating from the surface to the atmosphere B. water vapor
Shkiper50 [21]
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2 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
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