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mote1985 [20]
3 years ago
6

Early experiments with light were understood by explaining the behavior of light in terms of waves. Which experimental result re

quired considering the particle nature of light?
A. The ultraviolet catastrophe of blackbody radiation

B. The refraction of light through a prism

C. The interference of light traveling through thin films

D. The dark-and-light pattern seen with a diffraction grating
Physics
1 answer:
STALIN [3.7K]3 years ago
8 0
The answer to what experimental result required considering the particle nature of light is A. The ultraviolet catastrophe of blackbody radiation.
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A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
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6 0
2 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  

The initial angular momentum of the Raven is m_r * u_r * \frac{L}{2}

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}

            w = 1.43\  rad/sec

5 0
2 years ago
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