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3 years ago

The phase change in which a substance changes from a gas directly to a solid is

2 answers:

7 0

<span>Condensation is the change of the substance from liquid to solid phase. Example of this is the formation of ice. Vaporization is the change of substance from liquid to gas phase. Example of this is the boiling of water. Deposition is the change of a substance from gas to solid phase. Example of this is the formation of ice on a winter day. Sublimation is the change of a substance from solid to gas phase. Example of this is dry ice. The answer is letter C.</span>

ddd [48]3 years ago

5 0

Answer: Properties and States of Matter Practice

1. Which of the following are pure substances? Compounds

2. A substance that cannot be broken into similar substances is called a(n)... Element

3. When a physical change in a sample occurs, which of the following does NOT change? Composition

4. Which of the following will cause a decrease in gas pressure in a closed container? Lowering the temperature

5. The heat of fusion for water is the amount of energy needed for water to... Melt

6. The phase change in which a substance changes from a gas directly to a solid is... Deposition

7. Which of the following phase changes is an exothermic change? sublimation

8. Which of the following has the highest viscosity? Corn Syrup

9. In which of the substances in the figure below are the forces of attraction among the particles so weak that they can be ignored under ordinary conditions? It is the one where the particles are more apart.

Explanation:

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An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 25 m lower vert

Musya8 [376]

Answer:

x = 5.79 m

Explanation:

given,

mass of the car = 39000 Kg

spring constant = 5.7 x 10⁵ N/m

acceleration due to gravity = 9.8 m/s²

height of the track = 25 m

length of spring compressed = ?

using conservation of energy

potential energy is converted into spring energy

$m g h = \dfrac{1}{2}kx^2$

$x =\sqrt{\dfrac{2 m g h}{k}}$

$x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}$

$x =\sqrt{33.5263}$

x = 5.79 m

the spring is compressed to x = 5.79 m to stop the car.

3 0

3 years ago

A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en

dem82 [27]

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

= 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E = 180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

7 0

4 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

4 0

3 years ago

Even though you praise your dog for sitting inside on his bed and chewing the bone you gave him, your dog insists on going outsi

qwelly [4]

Answer:

dog

Explanation:

5 0

3 years ago

Final velocity will be greater than initial velocity of an object is

Natali [406]

Answer:

accelerating

Explanation:

If we consider(v > u) Acceleration:

final velocity(v)= 14m/s

initial velocity(u)=10m/s

time taken(t)= 2 seconds

a= $\frac{(v-u)}{t} =\frac{(14-10)}{2}$ =2m/s²

If we consider (v<u) Deceleration:

final velocity(v)= 3m/s

initial velocity(u)=9m/s

time taken(t)=2 seconds

a= $\frac{(v-u)}{t}=\frac{(3-9)}{2}$ = -3m/s²

4 0

3 years ago