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OLga [1]
3 years ago
12

Why does refracting light bend when it enters a denser or less dense medium?

Physics
1 answer:
Rzqust [24]3 years ago
4 0

Answer: The light bends because light travels fast but it slows down in a denser medium. For example light refracts in water or it bends after passing through air. When light passes through air ( a less dense medium ) then through water ( a more dense medium ) the beam of light bends because light travels more slowly in a denser medium then it picks up its pace again once it passes. The density of the substance determines how much the light is refracted. I hope this makes sense and I hope this answered your question!! :)

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A sinusoidal wave is traveling along a rope. The oscillator that generates the wave completes 40.0 vibrations in 30.0s . A given
Phantasy [73]

From the solution that I have done, the wavelength in the question that we have is 31.88 cm

<h3>How to solve for the wavelength</h3>

The frequency in the question is given as 40/30 = 1.33 hz

Next we have to solve for V

= 425/10

= 42.5 cm/s

v = frequency * wavelength

we have to put in the values in the formula. This would be

42.5 = 1.33 x wavelength

we have to divide through by 1.33 to get the wavelength. This would be

42.5/1.333 = wavelength

31.88 cm = wavelength

Hence we can say that the wavelength in the question that we have here is 31.88 cm

Read more on wavelength here:

brainly.com/question/10728818

#SPJ4

8 0
1 year ago
If the pendulum took longer to complete one oscillation, how would the graph change?
pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.


If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.


If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

line would be less.


If it was a graph of the pendulum's mass, weight, temperature, cost, etc.,

then the graph would be a horizontal line, and nothing that might change

the period of oscillation would have any effect on the graph.

7 0
2 years ago
Read 2 more answers
A geologist sees a fault along which blocks of rock in the footwall have moved higher relative to blocks of rock in the hanging
Luda [366]

Answer: Normal fault

Explanation:

The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

6 0
3 years ago
What is average speed?what is the formula for average speed?what does it mean?
Rama09 [41]
Average speed = total distance traveled/total time taken
it is the total distance traveled in a total time the total distance is attained
4 0
3 years ago
The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1
mr_godi [17]

Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m

Energy of one photon is given by the relation :

E=\frac{hc}{\lambda}    ....(1)

Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

E_{1}=N\times\frac{hc}{\lambda}

Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.

3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}

N = 1.35 x 10⁵

8 0
3 years ago
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