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poizon [28]
4 years ago
13

while tuning his violin the violinist compares the pitch his string produces to the pitch produced by an electronic tuner if the

tuner produces a note with a frequency of 440 hz and the violine plays a note with a frequency of 443 hz what is the beat frequency produced?
Physics
1 answer:
gayaneshka [121]4 years ago
4 0

Answer:

3 Hz

Explanation:

A beat is the pattern of interference among two waves with frequencies which are different.

The difference of the frequencies of the waves gives us the beat frequency.

Here, one frequency is f_1=440\ Hz, the other frequency is f_2=443\ Hz

So, the beat frequency

\Delta f=|f_1-f_2|\\\Rightarrow \Delta f=|440-443|\\\Rightarrow \Delta f=3\ Hz

The beat frequency that is produced is 3 Hz

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(NEED HELP BADLY)
spin [16.1K]

Answer:

The answer to your question is below

Explanation:

Data 1

mass 1 = 250

mass 2 = 250 kg

gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²

distance = 8 m

Formula

F = G\frac{m1m2}{r^{2} }

Substitution

F = 6.67 x 10^{-11} \frac{250 x 250}{8^{2} }

Result

F = 0.000000065 N

Data 2

mass 1 = 1000 kg

mass 2 = 1000 kg

distance = 5 m

Substitution

F = 6.67 x 10^{-11} \frac{1000 x 1000 }{5^{2} }

Result

F = 0.000002667 N      

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4 years ago
What is the speed of sound for noise that travels 2 km in 5.8
earnstyle [38]

Answer:

2.9

Explanation:

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3 years ago
A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the
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Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

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where

p_p is the initial momentum of the ping-poll ball

p_b is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

p'_p is the final momentum of the ping-poll ball

p'_f is the final momentum of the bowling ball

We can re-arrange the equation as follows

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or

-\Delta p_p = \Delta p_b

which means

|\Delta p_p | = |\Delta p_b| (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

I=\Delta p (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

|I_p| = |I_b|

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