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Sedaia [141]
3 years ago
15

What is the difference between flashing point, boiling point and melting point​

Physics
2 answers:
sp2606 [1]3 years ago
8 0
<h3><u>Answer and explanation;</u></h3>
  • <u>Melting point</u> is defined as the temperature at which solid and liquid phases are in equilibrium. It is the temperature at which a solid changes state from solid to liquid at atmospheric pressure.
  • <u>Boiling poin</u>t is the temperature at which the vapour pressure of a liquid is equal to the external pressure. It is the temperature at which a substance changes from a liquid into a gas.
  • <u>The flash point </u>of a flammable liquid or volatile liquid is the lowest temperature at which it can form an ignitable mixture in air. At this temperature the vapor may cease to burn when the source of ignition is removed.
Dvinal [7]3 years ago
4 0

Flashpoint mainly applies to volatile substances. It describes the lowest temperatures at which the vapors of the substance can ignite with the slightest spark. The melting point is the temperatures at which a substance changes phase from solid to liquid while the boiling point is the temperatures where a substance changes phase from liquid to gas.

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An object is 25.0 cm from the mirror, its height is 5.0 cm, and focal length is 8.0 cm. What is the distance from the image to t
Oksana_A [137]
1.) Use the formula to solve - 
             1/f = 1/do + 1/di; Where f = focal length; 1/do + 1/di
             1/f = 1/do + di
            1/8 = 1/25 + 1/?
          .125 = .04 + 1/di
   .125 -.04 = 1/di (transferred .04 to the left side of the equation)
       .085/1 = 1/di
.085di/.085 = 1/.085 (multiplied both sides by di and divided both sides by .085)
               di = 11.76 or 12
2.) Therefore, 12 cm is the distance from the image to the mirror
3 0
3 years ago
Read 2 more answers
Which of the following does not lead to addiction?
zloy xaker [14]

Answer:

Obesity does not lead to addiction. Obesity is a weight condition characterized by a body.

Explanation:

5 0
3 years ago
the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
3 years ago
Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis
WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{Mm}{r^{2}} (1)

Where:

F is the gravitational force between Earth and Moon

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant  

M=5.972(10)^{24} kg is the mass of the Earth

m=7.349(10)^{22} kg is the mass of the Moon

r=3.9(10)^{8} m is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to F:

F=m.a_{C} (2)

Where a_{C} is the centripetal acceleration given by:

a_{C}=\frac{V^{2}}{r} (3)  

Being V the orbital velocity of the moon

Making (1)=(2):

m.a_{C}=G\frac{Mm}{r^{2}} (4)

Simplifying:

a_{C}=G\frac{M}{r^{2}} (5)

Making (5)=(3):

\frac{V^{2}}{r}=G\frac{M}{r^{2}} (6)  

Finding V:

V=\sqrt{\frac{GM}{r}} (7)

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}} (8)

Finally:

V=1010.92 m/s

5 0
3 years ago
Why do electrons move from the negative end of the tube to the positive end
defon
Due to attraction ... of opposite charges
3 0
3 years ago
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