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3 years ago

What measurement can be used to determine the stability of the atmosphere?

1 answer:

5 0

I would say pressure, but it might depend on what you mean by stability. Pressure will tell you how many particles and how energized those particles are, the more particles the higher the pressure, the harder it is to change or remove the atmosphere due to the larger mass of it. Also consistent pressure measurements can tell you how stable the atmosphere in a particular region is. So if stability means how consistent are the weather conditions, pressure can be a good indicator too, differences and changing pressures can cause weather catastrophes like tornados and hurricanes.
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How many genders are there?

Dvinal [7]

Answer:2

Explanation:

8 0

3 years ago

Read 2 more answers

Titus drives his jetski a distance of 1000 meters in 7.045 seconds. How fast was he moving in meters per second? How fast was he

Ksenya-84 [330]

Answer:

a) v = 141.9 m/s

b) v = 317.4 miles/h

Explanation:

a) How fast was he moving in meters per second?

$v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s$

Hence, the jet ski is moving at 141.9 meters per second.

b) How fast was he moving in miles per hour?

$v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s*\frac{3600 s}{1 h}*\frac{1 mile}{1609.34 m} = 317.4 miles/h$

Therefore, the jet ski is moving at 317.4 miles per hour.

I hope it helps you!

5 0

2 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago

I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow

Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0

3 years ago

If yall friend me ill give lots of points like this in the future

Ne4ueva [31]

Answer:

ok......................

3 0

2 years ago