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Pachacha [2.7K]
2 years ago
9

The lowest point in Death Valley is 85.0 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m. What is

the change in gravitational potential energy of an energetic 65.0 kg hiker who makes it from the floor of Death Valley to the top of Mt. Whitney?
Physics
1 answer:
mestny [16]2 years ago
6 0

Answer:

The change in gravitational potential energy of the hiker = 2869685 J

Explanation:

Potential Energy: This is the energy possessed by a body, due to its change in position in the gravitational field. The unit of potential energy is Joules (J)

From the question,

Change in gravitational potential energy = Energy of the hiker at the top of  Mt. Whitney - Energy of the hiker at the floor of Death valley.

ΔEp = mgh₂ - mgh₁

ΔEp = mg(h₂-h₁)........................... Equation 1

Where ΔEp =  change in Potential Energy of the hiker, m = mass of the hiker, g = acceleration due to gravity, h₁ = lowest point in Death valley, h₂ = Elevation of Mt. Whitney.

Given: m = 65.0 kg, h₁ = -85 m ( because is a valley), h₂ = 4420 m,

Constant: g = 9.8 m/s²

Note: The h₁ is negative because is below sea level.

Substituting into equation 1

ΔEp = 65×9.8×[4420-(-85)]

ΔEp = 637(4420+85)

ΔEp = 637(4505)

ΔEp = 2869685

ΔEp = 2869685 J.

Thus the change in gravitational potential energy of the hiker = 2869685 J

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20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is n=\frac{c}{v}

26) Critical angle: [tex]sin \theta_c = \frac{n_2}{n_1}[/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

with

n_1, n_2 are the refractive index of 1st and 2nd medium

\theta_1, \theta_2 are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

  • A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
  • A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

  • A material with refractive index of 1.1
  • A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1 (1)

For light moving from air to water:

n_1 \sim 1.00 is the index of refraction of air

n_2 = 1.33 is the index of refraction ofwater

In this case, \frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75

For light moving from air to glass,

n_2 = 1.51 is the index of refraction of glass

And so

\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66

From eq.(1), we see that the angle of refraction \theta_2 is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

n=\frac{c}{v}

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

  • The higher the index of refraction, the slower the light
  • The lower the index of refraction, the faster the light

26)

From Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, n_1 > n_2, so \frac{n_1}{n_2}>1, and since sin \theta_2 cannot be larger than 1, there exists a maximum value of the angle of incidence \theta_c (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

sin \theta_c = \frac{n_2}{n_1}

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

sin \theta_c = \frac{n_2}{n_1}

For the interface glass-air,

n_1 \sim 1.51\\n_2 = 1.00

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}

For the interface glass-water,

n_1 \sim 1.51\\n_2 = 1.33

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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Answer:

Explanation:

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a_c=\frac{25^2}{200}

a_c=3.125\ m

(b)Force that provide centripetal acceleration

F=F_c=\frac{mv^2}{r}

F=\frac{1428.57\times 25^2}{200}

F=4464.285\ N

(c)Friction force between car and tires is given by

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Centripetal force will balance the friction force

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Answer:

t=0.704s

Explanation:

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Answer:

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