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Anon25 [30]
3 years ago
7

Calculate the index of refraction for a medium in which the speed of light is 1.239 ✕ 108 m/s. (The speed of light in vacuum is

2.998 ✕ 108 m/s. Enter your answer to at least three decimal places.)
Physics
1 answer:
True [87]3 years ago
3 0

Answer:

Refractive index, n = 2.419

Explanation:

It is given that,

Speed of light in the medium, v=1.239\times 10^8\ m/s

Speed of light in the vacuum, c=2.998\times 10^8\ m/s

Let n is the index of refraction for a medium. We know that the refractive index for any medium is equal to the ratio of speed of light in vacuum to the speed of light in medium. It can be written as :

n=\dfrac{c}{v}

n=\dfrac{2.998\times 10^8\ m/s}{1.239\times 10^8\ m/s}

n = 2.419

So, the index of refraction for a medium is 2.419. Hence, this is the required solution.

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3 years ago
During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a
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2 years ago
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains
OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

<u>C = 771.35 J/kg°C</u>

5 0
3 years ago
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

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4 0
3 years ago
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Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

3 0
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