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3 years ago

What happens in a cloud to cause electrons to move from some particles to others?

Physics

1 answer:

vova2212 [387]3 years ago

6 0

Answer:

Air molecules and suspended water droplets collide as they swirl around in the clouds. Warmer air and water droplets rise, carrying charges with them. The result is an excess of positive charge near the cloud tops, and an excess of negative charge in the bottom layers of the clouds.

they are knocked off some ice and added to other ice as they crash past each other

Explanation:

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wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg

Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg ( $M_J$ )

Speed of Jeremy is 3 m/s ( $V_J$ )

Speed of Jeremy after collision is ( $V_{JA}$ ) -2.5 m/s

Mass of Hans is 140 kg ( $M_H$ )

Speed of Hans is -2 m/s ( $V_H$ )

Speed of Hans after collision is ( $V_{HA}$ )

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is

= $=\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}$

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is

= $=\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}$

= 120 × (-2.5) + 140 × $V_{HA}$

= -300 + 140 × $V_{HA}$

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × $V_{HA}$

80 + 300 = 140 × $V_{HA}$

380 = 140 × $V_{HA}$

380/140= $V_{HA}$

$V_{HA}$ = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0

3 years ago

Research the potential for a large earthquake off the coast of Oregon. What are two of the biggest concerns with a large earthqu

Sphinxa [80]

There is a 33% chance of a large earthquake off the coast of Oregon.

The two main concerns of a large earthquake are Gas fires and Tsunami .

Explanation:

Everytime a large earthquake hits in any other part of the world, We find a small earthquake hitting off the coast of Oregon.

So scientists feel that as United States is prone to Earthquakes, the people have to be prepared for such eventualities.

The main concern for this large earthquake is, that it might lead to gas leakages at homes that can start a gas fire, which might spread. So people have to be careful and check on any gas leakages.

Such large earthquakes can also lead to tsunami waves. So people have to take precautions and evacuate, if they are near the coastline.

6 0

3 years ago

If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit

Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) = (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 = (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0

3 years ago

You drop a pencil from your desk, which is 1 meter above the floor. How long does it take for the pencil to hit the floor? How f

Oksanka [162]

Answer:

tardo50 ilimegnudos

Explanation:

8 0

3 years ago

The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A:

gizmo_the_mogwai [7]

Answer: B

Explanation:

Because the ramp is too high, the forces aren't enough to push onto the car stuck down.

6 0

3 years ago