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lana66690 [7]
3 years ago
10

A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in ai

r is 340 m/s. How many beats per second do the people on the fire engine hear
Physics
1 answer:
Svetach [21]3 years ago
8 0

Answer:

The  values is  f_b  =14.9 \  beats/s

Explanation:

From the question we are told that

   The  speed of the fire engine is  v =  5\ m/s

    The frequency of the tone is  f =  500 \ Hz

    The speed of sound in air is v_s  =  340 \ m/s

The  beat frequency is mathematically represented as

     f_b  =  f_a  -  f

Where  f_a is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as

   f_a  =  [\frac{v_s + v }{v_s  -v} ]* f

substituting values

  f_a  =  [\frac{340 + 5 }{340  -5} ]* 500

  f_a  = 514.9 \  Hz

Thus  

      f_b  =514.9 -  500

      f_b  =14.9 \  beats/s

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Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa
garik1379 [7]

Answer:

q =  400 nC

the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

            V = k q / r

they ask us for the electrical charge

          q = V r / k

let's calculate

         Q = 600 6.0 / 9 10⁹

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the correct answer is b

Traslate

La expresión para el potencial eléctrico de una carga puntual es

            V = k q/r

nos piden la carga eléctrica

          q= V r /k

          calculemos

         Q= 600  6,0 / 9 10⁹

         Q=  4 10⁻⁷ C

reduzcamos a nC

          Q = 4 10⁻⁷ C(10⁹ nC/1C )  

          q = 4 10² nC = 400 nC

la respuesta correcta es b

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3 years ago
How many electrons must be remowel from an electricaly Nurutral Silvor Coin to five it a charge of 3.2 NC ?
kakasveta [241]

Answer:

 #_electrons = 2 10¹⁰ electrons

Explanation:

For this exercise we can use a direct rule of three proportions rule. If an electron has a charge of 1.6 10⁻¹⁹ C how many electrons have a charge of 3.2 10⁻⁹ C

          #_electrons = 3.2 10⁻⁹ ( \frac{1}{1.6 \ 10^{-19}})

          #_electrons = 2 10¹⁰ electrons

7 0
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