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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

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.

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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Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.

Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1 : m1= 5kg

50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

F2 = 2*a

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

Look m1 free-body diagram:

∑Fx = m1*a

50-F1= 5 *a Equation 1

Look m2 free-body diagram:

∑Fx = m2*a

F1-F2= 3 *a Equation 2

Look m3 free-body diagram:

∑Fx = m3*a

F2 = 2*a Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

Look Free body diagram of the mass set

∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

(d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0

2 years ago

A force of 265.1 N acts on an object to produce an acceleration of 14.52 m/s^2. What is the mass of the object?

astra-53 [7]

The answer is :

18.26

Hope I helped.

6 0

3 years ago

Read 2 more answers

Using Newton's third law of motion, write a scientific explanation that describes why

Aneli [31]

Answer:

when a force is applied by one object to a second object, an equal and opposite force is applied back on the first object

Explanation:

4 0

1 year ago

An object experiences an impulse, moves and attains a momentum of 200 kg·m/s. If its mass is 50 kg, what is its velocity?

Nady [450]

Answer:

4 m/s

Explanation:

Momentum is defined as:

$p=mv$

where

m is the mass of the object

v is its velocity

For the object in this problem, we know:

p = 200 kg m/s is the momentum

m = 50 kg is the mass

Solving for the velocity, we find:

$v=\frac{p}{m}=\frac{200}{50}=4 m/s$

8 0

2 years ago

A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F

EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B = 1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ) eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0

3 years ago