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xenn [34]
3 years ago
12

Cellular phones use _____?

Physics
2 answers:
mina [271]3 years ago
6 0

3. D-Microwaves

1. C-electromagnetic spectrum

2. B-radio waves

^^(for those doing the three question physical science test)

dangina [55]3 years ago
3 0
Let's see: frequency of cellular phone waves (GSM phones) is (800-1900 MHz). If we look at the table of the electromagnetic spectrum, we can see that this range is contained within the frequencies of the microwaves, which include waves in the range 300 MHz-300 GHz.

So, summarizing, the correct answer is "microwaves".
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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

=>    \lambda_2 =  \frac{10  *   587 *10^{-9}}{11}

=>   \lambda_2 =  534 *10^{-9} \ m

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3 years ago
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Explanation:

Im sorry

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Complete the sentences below using the words below, you may use each word more than once
satela [25.4K]
Decreases, stays the same, increases. The volume decreases because as air is cooled, the individual molecules collectively possess less kinetic energy and the distances between them decrease, thus leading to a decrease in the volume they occupy at a certain pressure (please note that my answer only holds under constant pressure; air, as a gas, doesn't actually have a definite volume). The mass stays the same because physical processes do not create or destroy matter. The law of conservation of mass is obeyed. You're only cooling the air, not adding more air molecules. The density decreases because as the volume decreases and mass stays the same, you have the same mass occupying a smaller volume. Density is mass divided by volume, so as mass is held constant and volume decreases, density increases.
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