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rodikova [14]
3 years ago
6

You are watching a wave with a frequency of 512 Hz go past. How many peaks of the wave go past you each minute?

Physics
2 answers:
bija089 [108]3 years ago
6 0

Answer:

61440

Explanation:

The period of a sine wave is the length of one cycle of the wave and is characterized by having two peaks a negative and a positive (See the image).

The frequency on the waves is given in hz equivalent to one cycle per second, so 512 hz are 512 cycles per second, given that one minute is equivalent to 60 seconds we have a total of 60*512 = 30720 cycles per second, because each cycle contains two peaks the total amount of peaks in one minute is 30720 *2 =61440

inna [77]3 years ago
4 0

Answer:

61440 peaks

Explanation:

A hertz represents one cycle for every second:

f= Hz=\frac{1}{s}

So if a wave have frequency of 2Hz for example, this means the wave does two cycles per second.

Normally the waves are represented by cosine or sine waves. These kind of waves have two peaks in each cycle, one positive and one negative. With this in mind, let's calculate how many peaks of the wave pass each minute.

A minute has 60 seconds, hence:

60*512=30720\hspace{3}cycles\hspace{3}per\hspace{3}minute

And we know already that every cycle has two peaks, so:

30720*2=61440\hspace{3}peaks\hspace{3}per\hspace{3}minute

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Dibuja la gráfica de calentamiento de un kilogramo de plomo que se encuentra inicialmente a 70ºC y pasa a una temperatura final
MariettaO [177]

Answer:

Q= m c_e ΔT and   Q = m L

Explanation:

For this graph of temperature vs energy (heating) we must use two relations

* for when there is no change of state

          Q= m c_e ΔT

* for using there is change of state

          Q = m L

the second expression is a consequence of the fact that all the energy supplied is used to change the state of the solid-liquid and liquid-gas system

the energy supplied is the sum of the energy in each interval

divide the system into intervals determined by the state change points

1) from T₀ = 70ºC to T_f = 327.4ºC, sample in solid-liquid state

           c_e = 128 J / kg ºC

           Q₁ = m c_e (T_f -To)

           Q₁=1  128 (327.4 -70)

           Q₁ = 3.29 10⁴ J

           Q = Q₁ = 3.29 10⁴ J

2) when is it changing from solid to liquid

            L = 2.45 10⁴ J / kg

            Q2 = 1 2.45 10⁴

            Q2 = 2.45 10⁴ J

            Q = Q₁ + Q₂

             Q = 5.74 10⁴ J

3) from to = 327.4ºC until T_f = 1725ºC, sample in liquid state

in the tables the specific heat of the solid and liquid state is the same

             Q3 = m c_e (T_f -To)

             Q3 = 1 128 (1725 -327.4)

             Q3 = 1.79 10⁵ J

              Q = Q₁ + Q₂ + Q₃

              Q = (3.29 +2.45 + 17.9) 10⁴ J

              Q = 23.64 10⁴ J

4) for when it is changing from the liquid state to the gaseous state

             L_v = 8.70 10⁵ J / kg

             Q₄ = m L_v

             Q₄ = 1 8.70 10⁵

             Q₄ = 8.70 10⁵ J

             Q = Q₁ + Q₂ + Q₃ + Q₄

              Q = (3.29 +5.74 + 17.9+ 87.0) 10⁴ J

               Q = 110.64 10⁴ J

5) from To = 1725ºC to T_f = 2000ºC, sample in gaseous state

             Q₅ = m c_e ΔT

             Q₅ = 1 128 (2000 -1725)

             Q₅ = 3.52 10⁴ J

             Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

              Q = 114.16 104 J

the following table shows the points to be plotted

         Energy (10⁴ J)  Temperature (ºC)

                  0                     70

                 3.29             327.4

                 5.74             327.4

               23.64           1725

               110.64          1725

                114.16         2000

In the attachment we can see a graph of Temperature versus energy supplied

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