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nirvana33 [79]
3 years ago
9

If the average frequency emitted by a 160 W light bulb is 5.00 1014Hz and 10.0 of the input power is emitted as visible light ap

proximately how many visible light photons are emitted per second
Physics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

The value is \frac{n}{t}  = 4.83 *10^{19} \  photons / s

Explanation:

From the question we are told that

   The power rating of the bulb is  P = 160 \ W

   The frequency is f =  5.00 *10^{14} \ Hz

   The percentage of the input power that is emitted as visible light is \eta =  10\% = 0.10

   

Generally the amount of power emitted as visible light is mathematically represented as

       P_l =  0.10 * P_i

=>  P_l =  0.10 *160

=>  P_l =  16 \ W

Generally the amount of energy emitted as light is mathematically represented as

        E = n *  h  *  f

Here n is the number of photon ,  h is the Planks constant with value h =  6.625*10^{-34} \  J\cdot s

Generally this power emitted as visible light is mathematically represented as

   P_l = \frac{E}{t}

=>  P_l = \frac{E}{t} = \frac{nhf}{t}

=>  \frac{n}{t}  = \frac{P_l }{hf}

=>  \frac{n}{t}  = \frac{16 }{6.625 *10^{-34}* (5.00*10^{14})}

=>  \frac{n}{t}  = 4.83 *10^{19} \  photons / s

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Answer:

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Explanation:

Work done (W) is the product of the force (F) applied on a body and the distance (s) moved in the direction of the force.

i.e W = F × s

It is a scalar quantity and measured in Joules (J).

Given that: F = 35.0 N and s = 1.50 m, then;

W = F × s

W = 35.0 × 1.5

   = 52.5 J

Therefore, the work done on the sleigh by Samuel is 52.5 J.

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3 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
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Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

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A steel rod has a diameter of 3.75cm.Express it's diameter in millimeters and meter
romanna [79]

Answer:

3.75cm into millimetres

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Light enters an equilateral prism with an incident angle of 35° to the normal of the surface. Calculate the angle at which the
julia-pushkina [17]

Answer:

65.9°

Explanation:

When light goes through air to glass

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Sin r = 0.382

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Now the ray is incident on the glass surface.

A = r + r'

Where, r' be the angle of incidence at other surface

r' = 60° - 22.5° = 37.5°

Now use Snell's law at other surface

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Where, i' be the angle at which the light exit from other surface.

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vova2212 [387]

Answer:

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Explanation:

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