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SOVA2 [1]
3 years ago
5

Light travels at 3.0 × 108 m/s in a vacuum. Use the index of refraction for water to determine the speed of light in water. Roun

d your answer to the nearest tenth.
Physics
2 answers:
rusak2 [61]3 years ago
5 0
The answer is 2.3. I just answered this question and got it right.
sp2606 [1]3 years ago
3 0

<u>Answer:</u> The velocity of light in water is 225056264.1 m/s.

<u>Explanation:</u>

To calculate the speed of light in medium (water), we use the formula:

\eta=\frac{c}{v}

where,

\eta = index of refraction of water = 1.333

c = speed of light in vacuum = 3\times 10^8m/s

v = speed of light in water = ?m/s

Putting values in above equation, we get:

1.333=\frac{3\times 10^8}{v}\\\\v=2.2505626406\times 10^8m/s

To round the speed of light in water to the nearest tenth place, we will multiply the exponent with the answer and then round off the tenth place.

2.2505626406\times 10^8=225056264.06

Rounding 0.06 to 1 because 6 is more than 5, we get:

\Rightarrow 225056264.1m/s

Hence, the velocity of light in water is 225056264.1 m/s.

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3 years ago
The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard
gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

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8 0
3 years ago
A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

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at the highest point of the trajectory the vertical speed must be zero

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          a = (0i ^ - 9.8 j ^) m / s²

5 0
2 years ago
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