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Rainbow [258]
3 years ago
11

Explain what happens to a light wave that hits the surface of a pool.

Physics
1 answer:
brilliants [131]3 years ago
5 0

A light wave that hits the surface of a pool gets refracted and gives us an apparent image of the surface of the pool, following the concepts of refraction.

<u>Explanation:</u>

Let’s recall the concept of refraction when a light wave passes from medium of rarer to denser. There is a change in the speed of light while travelling from medium of rarer to denser.

There can be a change in the direction as well. This property is known as “Refraction” and the best example to see refraction is watching the surface of a clean pond, lake or pool.

When the light travels from a rarer medium (air) to a denser medium (water), it changes its angle of direction and gets refracted and hit to our eye lenses. With this, we see the surface of the pool at a changed angle and it seems to be a bit shallow than its original depth.

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What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?
igomit [66]

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

4 0
3 years ago
Read 2 more answers
A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
posledela

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

where

  • v represents the speed of this wave, and
  • \lambda represents the wavelength of this wave.

6 0
3 years ago
Explain the law of conservation of momentum.
WITCHER [35]

Answer:

It says that momentum cannot be created or destroyed. Momentum just passes through systems as opposed to just disappearing in thin air. Momentum can be changed to heat or energy.

Explanation:

3 0
3 years ago
A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief
fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

x_1 = -120cos70^o = -41.04 km

y_1 = -120sin70^o = -112.763 km

After the 2nd flight its coordinate would be:

x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km

y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km

So in order to fly back to its HQ it must fly a distance and direction of

s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km

tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71

\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o north of east

3 0
3 years ago
Nellie pulls on a 10kg wagon with a constant horizontal force of 30N. If there are no other horizontal forces, what is the wagon
s2008m [1.1K]

Answer:

The acceleration of the wagon is 3 m/s².

To calculate the acceleration of the wagon, we use the formula below.

Formula:

F = ma............. Equation 1

Where:

F = horizontal Force

m = mass of the wagon

a = acceleration of the wagon.

make a the subject of the equation

a = F/m.............. Equation 2

From the question,

Given:

F = 30 N

m = 10 kg

Substitute these values into equation 2

a = 30/10

a = 3 m/s²

Hence, the acceleration of the wagon is 3 m/s².

5 0
2 years ago
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