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nordsb [41]
3 years ago
8

A sound wave with a frequency of 400 Hz is moving through a solid object. If the wavelength of the sound wave is 8 m, what is th

e speed of sound traveling through the solid object?
50 m/s

3200 m/s

408 m/s

0.02 m/s
Physics
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

v=3200m/s

Explanation:

The velocity (speed) of a sound wave (any wave in fact) is related to its frequency and wavelength by the equation v=\lambda f. This is easy to remember since it's a distance divided by a time because frequency is the inverse of the period.

Since for sound traveling through the given solid object we have that f=400Hz and \lambda=8m, the speed of sound will be:

v=\lambda f=(8m)(400Hz)=3200m/s

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1) draw a simple circuit with a voltage source and four resistors wired in series
Norma-Jean [14]

Answer:

1)

In this circuit (see attachment #1), we have:

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- Four resistors: a resistor is a device which opposes to the flow of current. The property that describes by "how much" the resistor "opposes" to the flow of current is called "resistance", and it is indicated with the letter R.

- In this circuit, the 4 resistors are in series. Resistors are said to be in series when they are connected along the same branch of the circuit, so that the same current flow across each of them.

- For resistors in series, the equivalent resistance of the circuit is given by the sum of the individual resistances:

R=R_1+R_2+...+R_n

2)

In this circuit (see attachment #2), we have:

- A voltage source: as before, we have chosen a battery, providing an electromotive  force to the circuit

- Three resistors wired in parallel. Resistors are said to be connected in parallel when they are connected along different branches, but with their terminals connected to the same point, so that each of them has the same potential difference across it.

- For resistors in parallel, the equivalent resistance of the circuit is calculated using the formula:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}

3)

In this circuit (see attachment #3), we have:

- A voltage source (again, we have choosen a battery)

- Three resistors, of which:

-- 2 of them are connected in parallel with each other

-- the 3rd one it is in series with the first two

If we call R_1,R_2 the resistances of the first 2 resistors in parallel, their equivalent resistance is:

\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}\\\rightarrow R_{12}=\frac{R_1 R_2}{R_1+R_2}

Then, these two resistors are connected in series with resistor R_3; and so, the total resistance of this circuit will be:

R=R_{12}+R_3=\frac{R_1R_2}{R_1+R_2}+R_3=\frac{R_1R_2+R_3(R_1+R_2)}{R_1+R_2}

4)

In this circuit (see attachment #4), we have:

- A voltage source (again, a battery)

- We have 6 resistors, which are arranged as follows:

-- Two branches each containing 3 resistors

-- The two branches are in parallel with each other

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R_{123}=R_1+R_2+R_3

R_{456}=R_4+R_5+R_6

And since the two branches are in parallel, their total resistance will be:

\frac{1}{R}=\frac{1}{R_{123}}+\frac{1}{R_{456}}\\\rightarrow R=\frac{R_{123}R_{456}}{R_{123}+R_{456}}=\frac{(R_1+R_2+R_3)(R_4+R_5+R_6)}{R_1+R_2+R_3+R_4+R_5+R_6}

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