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nalin [4]
2 years ago
7

What name is given to a process where a substance melts under pressure?​

Physics
1 answer:
Digiron [165]2 years ago
8 0

Answer:

Melting

Explanation:

The process of a solid becoming a liquid is called melting. (an older term that you may see sometimes is fusion). The opposite process, a liquid becoming a solid, is called solidification. For any pure substance, the temperature at which melting occurs—known as the melting point—is a characteristic of that substance.

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When the heat source is removed from a fluid, convection currents in the fluid will eventually ____________then stop.
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When the heat source is removed from a fluid, convection currents in the fluid will eventually distribute heat uniformly throughout the fluid.  When all of the fluid is at the same temperature, convection currents will stop.

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3 years ago
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o
garri49 [273]

Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

  • molecular mass of glycerin, M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}
  • molecular mass of water, M_w=2+16=18\ g.mol^{-1}
  • ∵Density of water is 0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}
  • ∴mass of water in 316 mL, m_w=316\times 0.992=313.5 g
  • mass of glycerin, m_g=154\ g
  • pressure of mixture, P_x=55.32\ torr= 55.32\ mm.Hg
  • temperature of mixture, T_x=40^{\circ}C

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

n_g=\frac{m_g}{M_g}

n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

P_{sol}=50.4\ mm.Hg

7 0
3 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

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2 years ago
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