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Serhud [2]
3 years ago
8

Why is the evaporation of water a physical change and not a chemical change?

Physics
2 answers:
soldi70 [24.7K]3 years ago
6 0
The evaporation of water is a physical change because it isn't undergoing any chemical reactions. In other words nothing is being combined to form a new substance or substances.
Levart [38]3 years ago
5 0
When water turn into gas, it is still water molecules that are gaseous. If you boil a pot of water, the steam that is released is still water. It doesn't change to something else. A chemical change is a change where something completely changed. While a physical property is like a change of state ( solid, liquid or gas)
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The world’s longest organ pipe, in the boardwalk hall auditorium in atlantic city, is 64 feet long. what is the fundamental freq
zepelin [54]

The fundamental frequency of this open-open pipe is 8.82 Hz

The quantity of waves that pass a set location in a predetermined period of time is known as frequency. Frequency is the number of full cycles per second in the alternating current direction for an oscillating or fluctuating current. The hertz, also known as Hz, is the accepted unit of frequency.

The temporal rate of change observed in oscillatory and periodic phenomena, such as mechanical vibrations, audio signals (sound), radio waves, and light, is specified by the frequency, an essential parameter in science and engineering.

Assume vs = 344 m/s

f1 = vs/2L

   = 344 m/s/ 2∙64 ft/(3.281 ft/m)

   = 8.82 Hz

To know more about Frequency refer:

brainly.com/question/14131991

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If someone looks far enough into space, they should be able to see the beginning of the universe true of false
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The reaction is Exothermic

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A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
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