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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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The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 5.9 mm. Using

timurjin [86]

Answer:

45.88297 m

Violet

Explanation:

x = Gap between holes = 5.9 mm

$\lambda$ = Wavelength = 527 nm

D = Diameter of eye = 5 mm

L= Distance of observer from holes

From Rayleigh criteria we have the relation

$\frac{x}{L}=1.22\frac{\lambda}{D}\\\Rightarrow L=\frac{xD}{1.22\lambda}\\\Rightarrow L=\frac{5.9\times 10^{-3}\times 5\times 10^{-3}}{1.22\times 527\times 10^{-9}}\\\Rightarrow L=45.88297\ m$

A person could be 45.88297 m from the tile and still resolve the holes

Resolving them better means increasing the distance between the observer and the holes. It can be seen here that the distance is inversely proportional to the wavelength. Violet has a lower wavelength than red so, violet light would resolve the holes better.

5 0

3 years ago

A circuit has an overall resistance of 40 ohms and a current of 200mA.

svetoff [14.1K]

Answer:

Hey!

_________________

Voltage (V) = 0.8V

Current (I) = 200 mA = 200/10^3 = 2/10

Resistance = ?

Resistance = Voltage / Current

Voltage = Current × Resistance

0.8 = 2/10 × Resistance

0.8×10/2 = Resistance

8/2 = Resistance

Resistance = 4 ohm

_________________

Hope it helps...!!!

Explanation:

3 0

2 years ago

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'

kramer

Chameleon's tongue is more fast than thought. Its long sticky tongue moves at an amazing ballistic speed which lashes out unsuspecting insects and bugs. Now let us see how fast it is.

GIven:

acceleration of the chameleon's tongue- 260 m/s
2 for 20 ms
constant speed 30 ms
50 ms total time
1/20 of a second

solution:

<u>260</u> = <u> n</u><u> </u>
20 50
<u>20 n </u>= <u>13, 000</u>
20 20
n= 650 m/s

4 0

2 years ago

If an object emits or reflects light, A. it has a smooth surface. B. it can be seen by the eye. C. it must be the color white. D

kirill [66]

Answer:

It can be seen by the eye

Explanation:

Took the test

5 0

2 years ago

Read 2 more answers

A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall

Alexandra [31]

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=14.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .