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Goryan [66]
2 years ago
14

Which of the following is NOT a example

Physics
1 answer:
solmaris [256]2 years ago
5 0

Answer:

B. a piece of paper being torn

Explanation:

A chemical change is one that cannot be reversed. This means the original properties of the substance or object cannot be restored.

If you cook a raw egg, it would turn into a boiled egg (or a poached egg, however it is being cooked). The reaction is irreversible, so you cannot turn the cooked egg back into a raw egg - it is basically impossible to 'uncook' an already cooked egg.

When you toast a piece of bread, it turns into toast. You can't 'untoast' it back into bread. The chemical changes have already occurred and cannot be undone.

If you tear a piece of paper, it is still paper. You are only ripping it, not changing anything about it. You could simply tape the torn bit back to the original bit, or glue it - either way, it is still paper and nothing has occurred to drastically change the physical state of it.

Therefore, B is not a chemical change.

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If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? Wh
Fudgin [204]

The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

4 0
3 years ago
Two forces are going in opposite directions each force is 9.
professor190 [17]

Answer:

The net force is zero.

Explanation:

Two opposing and equal forces cancel each other out, giving you a net force of zero.

7 0
3 years ago
A ball is thrown horizontally from the top of a tall cliff. Neglecting air drag, what vertical distance will the ball have falle
adell [148]

The relevant equation to use here is:

y = v0 t + 0.5 g t^2

where y is the vertical distance, v0 is initial velocity = 0, t is time, g = 9.8 m/s^2

 

y = 0 + 0.5 * 9.8 * 3^2

<span>y = 44.1 meters</span>

5 0
3 years ago
One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe
Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

\mu = \frac{m}{l}

\mu = \frac{8.79*10^{-3}}{70*10^{-2}}

\mu = 0.01255kg/m

The expression for the wavelength of the standing wave for the second overtone is

\lambda = \frac{2}{3} l

Replacing we have

\lambda = \frac{2}{3} (70*10^{-2})

\lambda = 0.466m

The frequency of the sound wave is

f_s = \frac{v}{\lambda_s}

f_s = \frac{344}{0.768}

f_s = 448Hz

Now the velocity of the wave would be

v = f_s \lambda

v = (448)(0.466)

v = 208.768m/s

The expression that relates the velocity of the wave, tension on the string and linear mass density is

v = \sqrt{\frac{T}{\mu}}

v^2 = \frac{T}{\mu}

T= \mu v^2

T = (0.01255kg/m)(208.768m/s)^2

T = 547N

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the n^{th} harmonic frequency is

f_n = nf_1

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

n=3

Then,

f_3 = 3f_1

Rearranging to find the fundamental frequency

f_1 = \frac{f_3}{3}

f_1 = \frac{448Hz}{3}

f_1 = 149.9Hz

7 0
3 years ago
What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of n
Yuri [45]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  wavelength is  \lambda  =  622 nm

Explanation:

  From the question we are told that

    The distance of the slit to the screen is  D  = 5 \ m

    The order of the fringe is m  =  6

     The distance between the slit is  d = 0.9 \ mm  =  0.9 *10^{-3} \ m

    The fringe distance is  Y =  1.9 \ cm  =  0.019 \ m

Generally the for a dark fringe the fringe distance is  mathematically represented as

        Y  = \frac{[2m  - 1 ] *  \lambda *  D  }{2d}

=>     \lambda  =  \frac{Y *  2 *  d }{[2*m  -  1] *  D}

substituting values

=>      \lambda  =  \frac{0.019 *  2 *  0.9*10^{-3} }{[2*6  -  1] *  5}

=>     \lambda  =  6.22 *10^{-7} \ m

       \lambda  =  622 nm

8 0
3 years ago
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