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3 years ago

What must happen for liquid water to become water vapor?

Physics

2 answers:

motikmotik3 years ago

8 0

The water molecules must gain kinetic energy in order to become water vapour

il63 [147K]3 years ago

6 0

Answer:

<h2>C. The water molecules must gain kinetic energy.</h2>

Explanation:

When liquid water become water vapor, molecules move more, because they force between them is less strong. Notice that vapor water as lower consistency, it's like smoke where molecules can way anywhere, at this stage the force between molecules is not strong enough to maintain a ridig shape.

Now, at vapor stage, molecues have to gain more kinetic energy, because the least is the force between molecules, the more kinetic energy is gonna there.

Therefore, the right answer is C.

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In a classroom demonstration, students are using a Slinky to observe and learn about wave properties. If the Slinky has a period

Andrej [43]

Answer:

Frequency = 3.0 Hertz

Explanation:

Given the following data;

Period = 0.333 seconds

To find the frequency;

Mathematically, frequency of a wave is given by the formula;

Frequency = 1/period

Substituting into the formula, we have;

Frequency = 1/0.333

Frequency = 3.0 Hertz

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2 years ago

Determine the minimum size of glass tubing that can be used to measure water level. If the capillary rise in the tube does not e

Marina CMI [18]

Answer:

Explanation:

The formula to determine the size of a capillary tube is

h = 2•T•Cos θ / r•ρ•g

Where

h = height of liquid level

T = surface tension

r = radius of capillary tube

ρ = density of liquid

θ = angle of contact = 0°

g =acceleration due to gravity=9.81m/s²

The liquid is water then,

ρ = 1000 kg / m³

Given that,

T = 0.0735 N/m

h = 0.25mm = 0.25 × 10^-3m

Then,

r = 2•T•Cos θ / h•ρ•g

r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81

r = 5.99 × 10^-3m

Then, r ≈ 6mm

The radius of the capillary tube is 6mm

So, the minimum size is

Volume = πr²h

Volume = π × 6² × 0.25

V = 2.83 mm³

The minimum size of the capillary tube is 2.83mm³

8 0

2 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

Which of the following is NOT an exercise myth?

Blababa [14]

It is <span>C. Low to moderate level of exertion can be sustained over long periods of time </span>

7 0

2 years ago

YOU WILL GET BRAINLIEST Geologists can determine the relative ages of sedimentary layers by using the Law of Superposition. This

Lilit [14]

I personally think the answer is B.
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Hoped I helped!

8 0

2 years ago

Read 2 more answers