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postnew [5]
3 years ago
8

When light travels through a small hole, it appears to be an observer that the light spreads out, blurring the outline of the ho

le. Does this observation support the theory of light as a wave, or light being made of particles? Explain.
Physics
1 answer:
3241004551 [841]3 years ago
7 0

Answer:

support lights as a wave

Explanation:

In the model of light as a particle, the experimenter would expect to see one small hole of light emerging on the wall. However, as the light spreads out, it behaves much like a wave that diffracts when going through a small hole.

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When the dust is too thick to penetrate with visible light, such as the Nebula, Radio Waves are used to penetrate the dust. Longer radio waves can completely penetrate the thick cloud cover, allowing scientists to beam radar waves.
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3 years ago
Does heat demagnetize?
Nastasia [14]

Answer:

Yes

Explanation:

Heat affects the magnets because it confuses and misaligns the magnetic domains, causing magnetism to decrease

3 0
3 years ago
A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu
IrinaK [193]
The centripetal force is:
 F = mv² / R
 Where:
 m: mass of the object
 v: object speed
 R: radius of the curve.
 We have to:
 m = 2000kg
 v = 25 m / s
 R = 80 meters.
 Then the centripetal force acting on the vehicle is:
 F = (2000kg * (25m / s) ²) / 80m
 F = 15625 N
4 0
3 years ago
A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor
Kaylis [27]
<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span>U = R*i → i =  \frac{U}{R}<span>

Solving:

</span>i = \frac{U}{R}
i = \frac{12}{5}
\boxed{i = 2.4A}

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>
5 0
3 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
3 years ago
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