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algol [13]
3 years ago
5

The wavelengths of light emitted by a firefly span the visible spectrum but have maximum intensity near 550 nm. A typical flash

lasts for 100 ms and has a power of 1.2 mW. If we assume that all of the light is emitted at the peak-intensity wavelength of 550 nm, how many photons are emitted in one flash?
Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

3.3\cdot 10^{15}

Explanation:

First of all, let's calculate the energy of a single photon of wavelength

\lambda=550 nm=5.5\cdot 10^{-7}m

which is given by

E_1 = \frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.5\cdot 10^{-7} m}=3.6\cdot 10^{-19} J

The power of the flash is

P=1.2 mW=0.0012 W

and the time it lasts is

t=100 ms=0.1 s

so the total energy delivered in one flash is

E=Pt=(0.0012 W)(0.1 s)=1.2\cdot 10^{-3}J

This energy contains exactly N photons each of energy E_1, so the number of photons emitted in one flash is

N=\frac{E}{E_1}=\frac{1.2\cdot 10^{-3} W}{3.6\cdot 10^{-19}J}=3.3\cdot 10^{15}

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Answer:

25032.47 W

Explanation:

Power is the time rate of doing work, hence,

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Work done (non conservative) is given as:

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Total K. E. = 0.5mv²- 0.5mu²

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=> W = 0.5mv² - mgh(f)

P = [0.5mv² - mgh(f)] / t

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3 years ago
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A diode for which the forward voltage drop is 0.7 V at 1.0 mA is operated at 0.5 V. What is the value of the current
Vesnalui [34]

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Explanation:

The computation of the value of the current is given below:

z_i = I_s e^{\frac{0.7}{ut} }= 10^{-3}\\\\Z_z = I_s e^{\frac{0.5}{ut} }\\\\\frac{Z_z}{Z_i}= \frac{Z_z}{10^{-3}}  = e^{\frac{0.5\times 0.7}{0.025} }\\\\= 0.335 \mu A

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3 years ago
A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its
True [87]

Answer:

The angular velocity is  w = 1.43\  rad/sec

Explanation:

From the question we are told that

   The  mass of wooden gate  is m_g = 4.5 kg

    The  length of side is  L = 2 m

    The mass of the raven is  m_r = 1.2 kg

     The initial speed of the raven is u_r = 5.0m/s

     The final speed of the raven is   v_r = 1.5 m/s

From the law of  conservation of angular momentum we express this question mathematically as

       Total initial angular momentum  of both the Raven and  the Gate =  The Final angular momentum of both the Raven and the Gate  

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Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is  zero

Note: This above is the generally formula for angular momentum of  square objects

  The final angular velocity  of the Raven is  m_r * v_r * \frac{L}{2}

   The  final angular velocity of the Gate  is   \frac{1}{3} m_g L^2 w

Substituting this formula

  m_r * u_r * \frac{L}{2}  =   \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r * v_r * \frac{L}{2} -   m_r * u_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r *  \frac{L}{2} * [u_r - v_r]

Where w is the angular velocity

     Substituting value  

   \frac{1}{3} (4.5)(2)^2  w   =    1.2 *  \frac{2}{2} * [5 - 1.5]

     6w = 4.2

       w = \frac{6}{4.2}

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