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GenaCL600 [577]

2 years ago

7

The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed

before sound reaches the listener. (Speed of sound in air is 340 m/s.)

1 answer:

dmitriy555 [2]2 years ago

4 0

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

Speed of light, v = 340 m/s

Frequency, f = ?

we know:

wave number = $\frac{frequency}{speed of light}$

$20 = \frac{f}{340 m/s} \\\\f = 20 X 340 s^-^1\\\\f = 6800 Hz$

Therefore, the frequency of the wave is 6800 Hz

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a) The student hypothesizes that a greater fraction of kinetic energy is lost from the system during the collision when the spee

Mashcka [7]

Answer with Explanation:

One hypothesis above would probably rely on the situation called <em>"inelastic collision." </em>The two carts are travelling towards each other at different speed. So, this means that once they collide, the kinetic energy will not be conserved but will be transferred as <em>sound energy, thermal energy or material deformation</em>. The impact will be great, thus a <u>greater fraction of kinetic energy will be lost from the system.</u>

6 0

2 years ago

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

3 years ago

Need help on this please

Ratling [72]

question 2 answer is ALL OF THE ABOVE

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8 0

2 years ago

The __________ states that the firing rate of neurons in the auditory nerve matches the frequency of the sound wave.

borishaifa [10]

Correct answer choice is:

C. Volley principle (FREQUENCY MATCHING)

Explanation:

Volley theory declares that groups of neurons of the hearing rule counter to a noise by firing action potentials imperceptibly out of the stage with one another so that when connected, a higher pulse of sound can be encoded and transmitted to the brain to be examined.

7 0

3 years ago

Read 2 more answers