All categories

3 years ago

What kind of change is it when a liquid turns into a gas by boiling????????????????????

1 answer:

5 0

It is called vaporization. Vaporization is the phase transition from a liquid to a gas by means of evaporation or boiling. Evaporation occurs at temps below the boiling point and occurs on the liquids surface. Boiling is a rapid vaporization that occurs above the boiling temp and below or at the liquids surface.

You might be interested in

Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.

lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects Mass Force

1 10 kg 4 N

2 100 grams 20 N

3 10 grams 4 N

4 1 kg 20 N

Acceleration of object 1,

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2$

Acceleration of object 2,

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2$

Acceleration of object 3,

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2$

Acceleration of object 4,

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2$

It is clear that the acceleration of object 3 is $400\ m/s^2$ and it is greatest of all. So, the correct option is (3).

4 0

2 years ago

Read 2 more answers

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re

valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µ n

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

4 0

2 years ago

A transformer is designed to convert the voltage from 240 V to 12 V, and has 4000 turns on its primary coil. The turns on its se

harkovskaia [24]

Answer:

yaayyaysysysa

Explanation:

gsgssggsgsgsysyshehsh

4 0

2 years ago

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by stri

gogolik [260]

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

$K = 0.5 mv^2$

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2

(b) A the distance s same so the final velocities are also same.

8 0

2 years ago

two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of m

alexdok [17]

The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.

F α Qq

If the charges are now half the initial charges:

F α (1/2)Q *(1/2)q

F α (1/4)Qq

The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.

4 0

3 years ago

Read 2 more answers