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Degger [83]
3 years ago
8

Assuming that 10.0% of a 100 W light bulb's energy output is in the visible range (typical for incandescent bulbs) with an avera

ge wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away (in km) would you be if 468 photons per second enter the 3.30 mm diameter pupil of your eye
Physics
1 answer:
Verizon [17]3 years ago
5 0

Answer:

Explanation:

10% of 100 W = 10 W .

energy of each photon = hν , h is plank's constant and ν is frequency

= hc / λ , c is velocity of light , λ is wavelength

wavelength given λ  = 580 x 10⁻⁹ m .

energy of each photon = 6.6 x 10⁻³⁴ x 3 x 10⁸ / 580 x 10⁻⁹

= .034137 x ⁻¹⁷ J

no of photons in 10 W

= 10 /  .034137 x ⁻¹⁷

= 292.9 x 10¹⁷

If d be the required distance

no of photons per unit area per second  at distance d

= 292.9 x 10¹⁷ / 4πd²

no of photons through the area of eye

= π x (1.65 x 10⁻³)² x 292.9 x 10¹⁷ / 4πd²

= 199.355 x 10¹¹ / d²

199.355 x 10¹¹ / d² = 468 ( given )

d² = 199.355 x 10¹¹ / 468

= .42597 x 10¹¹

= 42597 x 10⁶

d = 206.4 x 10³ m

= 206.3 km.

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λ = 2 m.

The easiest way to solve this problem is using the equation of frecuency of a wave f = v/λ, where v is the velocity of the wave, and λ is the wavelength.

To calculate the wavelength of a microwave light travels through a liquid, it moves at a speed of 2.2 x 10⁸ m/s. If the frecuency of the light wave is 1.1 x 10⁸ Hz, we have to clear λ from the equation f = v/λ:

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Explanation:

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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
Anni [7]

Answer:

0.074m/s

Explanation:

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m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Where,

m_1 = mass of ball

m_2 = mass of the person

u_1 = Velocity of ball before collision

u_2 = Velocity of the person before collision

v_1 = velocity of ball afer collision

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We know that after the collision, as the person as the ball have both the same velocity, then,

v_1 = v_2

m_1u_1 + m_2u_2 = (m_1+m_2)v_2

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v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}

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m_1= 0.425kg

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u_2= 0m/s

Substituting,

v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}

v_2 = 0.074m/s

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

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3 years ago
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Genrish500 [490]

Answer:

No

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