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Semmy [17]
3 years ago
6

Which best explains the relationship between evaporation and temperature?

Physics
2 answers:
kogti [31]3 years ago
8 0
D) a liquid evaporates faster at higher temperatures
Svet_ta [14]3 years ago
4 0

Answer: Option (d) is the correct answer.

Explanation:

Evaporation is a process in which liquid phase of a substance changes into vapor or gaseous phase.

In a liquid, molecules are less tightly held together as compared to solids. Therefore, molecules of a liquid are able to slide past each other. But when a liquid is heated at a high temperature then molecules gain more kinetic energy.

As a result, number of collision increases between the molecules and hence they start to evaporate from the liquid.

Thus, we can conclude that the statement a liquid evaporates faster at higher temperatures because more particles have a higher speed and can overcome attractions in the liquid, best explains the relationship between evaporation and temperature.

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A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:
Igoryamba

Answer:

9.8\; {\rm J}, assuming that the gravitational field strength is g = 9.8\; {\rm N \cdot kg^{-1}}.

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be 0. External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let m denote the mass of this block. It is given that m = 1\; {\rm kg}. The weight of this block would be:

\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}.

Hence, the force that the boy applies on this block would be upward with a magnitude of F = 9.8\; {\rm N}.

The mechanical work that a force did is equal to the product of:

  • the magnitude of the force, and
  • the displacement of the object in the direction of the force.

The displacement of this block (upward by s = 1\; {\rm m}) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:

  • the magnitude of the force that this boy exerted, F = 9.8\; {\rm N}, and
  • the displacement of this block in the direction, s = 1\; {\rm m}.

\begin{aligned}\text{work} &= F\, s \\ &= 9.8\; {\rm N} \times 1\; {\rm m} \\ &= 9.8\; {\rm J}\end{aligned}.

5 0
2 years ago
Solve the following system of equations by using either substitution or elimination.
strojnjashka [21]
I believe that the answer to the question provided above are the following;

x = 29.8410

y = 16.6794

z = -1.2642
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
5 0
3 years ago
Why is damage from sound waves is an issue on the launchpad but not in the air
elixir [45]
Because it is if you know you know and it is also helping the sentwcnde and air and confiscation
6 0
2 years ago
A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)
MrRissso [65]

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

4 0
3 years ago
An object is placed in front of a convex mirror with a radius of curvature of magnitude 10 cm. The mirror produces an image that
jek_recluse [69]

Answer:

u = - 20 cm

m =\frac{1}{5}

Given:

Radius of curvature, R = 10 cm

image distance, v = 4 cm

Solution:

Focal length of the convex mirror, f:

f = \frac{R}{2} = \frac{10}{2} = 5 cm

Using Lens' maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Substitute the given values in the above formula:

\frac{1}{5} = \frac{1}{u} + \frac{1}{4}

\frac{1}{u} = \frac{1}{5} - \frac{1}{4}

u = - 20 cm

where

u = object distance

Now, magnification is the ratio of image distance to the object distance:

magnification, m =\frac{|v|}{|u|}

magnification, m =\frac{|4|}{|-20|}

m =\frac{4}{20}

m =\frac{1}{5}

4 0
3 years ago
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