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marissa [1.9K]
3 years ago
5

Can something change states of matter how do you know

Physics
1 answer:
belka [17]3 years ago
4 0
Yes, it can change states of matter from heat. Like ice, ice changes it’s state of matter from the temperature it is in, which causes it to turn into a liquid.(another form of matter) Again with heat, if water is boiled to a certain temperature it converts to a gas. Now with cold. If water is at such a coke temperature it will turn to ice, a form of matter. Now when a gas is cooled down, the atoms comprising the gas have less energy to move around.
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Every magnet must have what at its ends?
slega [8]

Explanation:

The correct is A.

every magnet have two pole

4 0
2 years ago
Using our understanding of the law of gravity. What happens to the gravity as we triple the distance between two objects?
MA_775_DIABLO [31]

Answer:

1/9

Explanation:

<em>Newton’s Law of Universal Gravitation </em>

Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.

F = GMm/r²

where  

F - the gravitatioal force in Newtons,  

M   and m  -two masses in kilograms  

r  - the separation in meters.  

G  - the gravitational constant (6.674*10 ⁻¹¹ N (m/kg) ² )

Because of the magnitude of  G , gravitational force is very small unless large masses are involved.

So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance

F1 ∝ 1/r² ---------------(1)

F2 ∝ 1(3r)²

F2 ∝ 1/9r²--------------(2)

(2)/(1)

\frac{F_2}{F_1} =\frac{1}{9}\\ F_2 =\frac{F_1}{9}

From their you get as the distance tripled, Force reduce by a factor of 9(3³)

for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)

3 0
3 years ago
When the pressure on a gas increases,what does the volume do?
kupik [55]

Answer:

As pressure goes up, volume goes down.

Explanation:

Pressure and volume of a gas are inversely proportional.  This means that as pressure goes up, volume goes down.  And as volume goes up, pressure goes down.

Cheers.

4 0
3 years ago
Read 2 more answers
Why are different constellations<br> of stars seen during different<br> seasons?
slamgirl [31]
Actually, they're not.  There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around.  And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night. 

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ?  Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ?  Now they're straight in the
direction of the sun.  So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.
5 0
3 years ago
Read 2 more answers
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
2 years ago
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