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ra1l [238]
2 years ago
5

A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave,

in kHz
Physics
1 answer:
Oliga [24]2 years ago
6 0

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

<u>Given the following data:</u>

  • Period = 645 μs

Note: μs represents microseconds.

<u>Conversion:</u>

1 μs = 1 × 10^-6 seconds

645 μs = 645 × 10^-6 seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

Frequency = \frac{1}{Period}

Substituting the value into the formula, we have;

Frequency = \frac{1}{645 * 10^-6}

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

<u>Conversion:</u>

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = 0.001 × 1550.39

X = 155039 kHz

To 3 significant figures;

<em>Frequency = 155 kHz</em>

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

Find more information: brainly.com/question/23460034

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Answer:

The well is 23.3 m

Explanation:

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A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o
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For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

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3 years ago
Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i
Lina20 [59]

Answer:

214.9 \Omega

Explanation:

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V=IR=(0.155 A)(38.7 \Omega)=6.0 V

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega

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3 years ago
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Answer:

<h3>true</h3>

Explanation:

<h3>hope it helps you ❤️</h3><h3>happy to help</h3>
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