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blsea [12.9K]
3 years ago
12

What do we call changes between solid liquid and gaseous forms of a substance

Physics
2 answers:
Nadusha1986 [10]3 years ago
6 0

Answer:

a change of state

Explanation:

const2013 [10]3 years ago
5 0

<u>We call changes between solid liquid and gaseous forms of a substance as  phase change or change of state.</u>

<u>Explanation:</u>

To change a substance from one state to another, extreme temperatures or pressures are required. Sometimes when a substance doesn't change states we should use all the ideas when that happens. To create a solid, we should decrease the temperature by a huge amount and then add pressure. For example, oxygen will solidify at -361.8 degrees Fahrenheit at standard pressure. However, it will freeze at warmer temperatures when the pressure is increased.

Phase changes happen when a substance reach some special points. Sometimes when a liquid becomes a solid a freezing point or melting point is used to measure the temperature at which a liquid changes into a solid. Some of the phase changes are: Condensation, Freezing, Melting.

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Calculate the molecular weight of Aluminium hydroxide​
Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

<h3><em>Al(OH)3 = 78.01 g/mol</em></h3>
5 0
3 years ago
A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     v_{y} = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

7 0
3 years ago
If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W
Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

P=\frac{3600 J}{3.0 s}=1200 W

5 0
3 years ago
Read 2 more answers
Is the earth's gravitational force on the sun larger than, smaller than, or equal to the sun's gravitational force on the earth?
Leona [35]

Answer:

The earth's gravitational force on the sun is equal to the sun's gravitational force on the earth

Explanation:

Newton's third law (law of action-reaction) states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In other words, when two objects exert a force on each other, then the magnitude of the two forces is the same (while the directions are opposite).

In this problem, we can call the Sun as "object A" and the Earth as "object B". According to Newton's third law, therefore, we can say that the gravitational force that the Earth exerts on the Sun is equal (in magnitude, and opposite in direction) to the gravitational force that the Sun exerts on the Earth.

6 0
3 years ago
A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected
gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0
3 years ago
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