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2 years ago

What is the period of a sound wave having a frequency of 340 Hz

Physics

2 answers:

sertanlavr [38]2 years ago

4 0

Answer:

lambda = 343 m/s divided by 340 Hz = 1.009 seconds

Hope it helps and have a wonderful day!

algol [13]2 years ago

4 0

Answer:

2.9x10^-3 secs

Explanation:

Data obtained from the question include:

f (frequency) = 340 Hz

T (period) =?

The period is simply the time taken to make one complete oscillation or vibration. The period is commonly referred to as the inverse of frequency.

T = 1/f

T = 1/340

T = 2.9x10^-3 secs

The period of the wave is 2.9x10^-3 secs

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Answer:

A 70 kg box is slid along the floor by a 400 n force. The coefficient of friction between the box and the floor is 0. 50 when the box is sliding

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2 years ago

Why x ray is called an electromagnetic wave<br>

Dahasolnce [82]

Because they behave just like all the electromagnetic waves of the spectrum. Same equations, just shorter wavelengths and more energy.

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2 years ago

Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is t

NikAS [45]

Answer:

f = 632 Hz

Explanation:

As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength

here we know that

$\Delta x = (2n + 1)\frac{\lambda}{2}$

given that path difference from two loud speakers is given as

$\Delta x = 5.80 m - 3.90 m$

$\Delta x = 1.90 m$

now we know that it will have fourth lowest frequency at which destructive interference will occurs

so here we have

$\Delta x = 1.90 = \frac{7\lambda}{2}$

$\lambda = \frac{2 \times 1.90}{7}$

$\lambda = 0.54 m$

now for frequency we know that

$f = \frac{v}{\lambda}$

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3 years ago

In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:

vesna_86 [32]

Answer:

D. 1×10⁻⁸ M

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2 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

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Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

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The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

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Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

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2 years ago