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1 year ago

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a 1020-hertz sound wave travels at 340 m/s in air with a wavelength of a) 30 m. b) 3 m. c) 0.333 m. d) 1 m. e) none of the above

choices are correct.

1 answer:

eimsori [14]1 year ago

5 0

The wavelength of the sound wave is equal to 0.333 m. Therefore, option (c) is correct.

<h3>What are frequency and wavelength?</h3>

The frequency of the wave can be defined as the number of oscillations that occur in one second and can be expressed in hertz. The wavelength can be defined as the distance between the two adjacent points of a wave such as two crests or troughs.

The relationship between frequency (ν), speed of sound waves (V), and wavelength (λ):

V = νλ

Given, the frequency of the sound wave, ν = 10 Hz

The speed of the sound wave, $V = 340 m/s$

The wavelength of the sound waves can determine as follows

λ = V/ν = 340/1020 = 0.333 m.

Therefore, the wavelength of the sound wave is 0.333 m

Learn more about wavelength and frequency, here:

brainly.com/question/18651058

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3 years ago

In an elastic collision, a 300 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

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Answer:

If we had:

$v_{1i}=5.3m/s$

$v_{2i}=5.9m/s$

We will have:

$v_{1f}=5.9m/s$

$v_{2f}=5.3m/s$

Explanation:

In an elastic collision both linear momentum and kinetic energy are conserved, so we will have:

$p_i=p_f$

$K_i=K_f$

We will call our bumpers 1 and 2.

For the momentum equation we know that:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Since all the masses are the same (300kg), they cancel out:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

For the kinetic energy equation we know that:

$\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}$

Since all the masses are the same (300kg), they cancel out (and also the 2 dividing):

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

We then must solve this system:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

Which we will rewrite as:

$v_{1i}-v_{1f}=v_{2f}-v_{2i}$

$v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2$

The last of these equations can be written as:

$(v_{1i}+v_{1f})(v_{1i}-v_{1f})=(v_{2f}+v_{2i})(v_{2f}-v_{2i})$

But we know that $v_{1i}-v_{1f}=v_{2f}-v_{2i}$ , so those cancel out:

$v_{1i}+v_{1f}=v_{2f}+v_{2i}$

So we can write:

$v_{1i}-v_{1f}+v_{2i}=v_{2f}$

$v_{1i}+v_{1f}-v_{2i}=v_{2f}$

Which means:

$v_{1i}-v_{1f}+v_{2i}=v_{1i}+v_{1f}-v_{2i}$

Which solving for the final velocity leaves us with:

$v_{2i}+v_{2i}=+v_{1f}+v_{1f}$

$v_{1f}=v_{2i}$

Grabbing any equation that relates both final velocities easily, for example $v_{1i}-v_{1f}+v_{2i}=v_{2f}$ , we obtain:

$v_{2f}=v_{1i}-v_{1f}+v_{2i}=v_{1i}-v_{1f}+v_{1f}=v_{1i}$

So we conclude that the bumpers have just exchanged velocities (something sometimes seen in billiards for example):

$v_{1f}=v_{2i}=5.9m/s$

$v_{2f}=v_{1i}=5.3m/s$

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3 years ago