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devlian [24]
2 years ago
14

A string of length 10.0 m is tied between two posts and plucked. This sends a wave down the string moving at a speed of 130 m/s

with a frequency of 215 Hz. How many complete wavelengths of this wave will fit on the string?
Physics
1 answer:
soldi70 [24.7K]2 years ago
8 0

Answer:

16.

Explanation:

  • In any wave, by definition, there exists a fixed relationship between the speed v, the frequency f , and the wavelength λ, as follows:

        v = \lambda * f (1)

  • In our case, v = 130 m/s and f= 215 Hz, so solving for λ in (1), we get:

        \lambda = \frac{v}{f} = \frac{130m/s}{215 hz} = 0.61 m (2)

  • In order to know how many wavelengths of this wave will fit on the string, we need just do divide the length of the string (10.0 m) over one single wavelength, as follows:

       n = \frac{L}{\lambda} = \frac{10.0m}{0.61m} = 16.4 (3)

  • Since we need to take the integer value from this expression, the number of complete wavelengths that will fit on this string is just 16.
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Imagine that I have a ping-pong ball and a bowling ball resting on the floor of our classroom. I go up to the bowling ball and g
MatroZZZ [7]

Answer:

the speed and aceleration of the ping pong ball is greater than that of the bowling ball.

Explanation:

We can analyze this exercise from several points of view, if we use Newton's second law

Bowling ball

           F = M a₁

pingpongg ball

           F = m a₂

as the forces the same

          M a₁ = m a₂

          a₂ = \frac{M}{m} a₁

Since the mass of the bowling ball is much greater than the ping pong ball,

          a₂ »a₁

so the acceleration of the ping pong ball is much greater than the acceleration of the bowling ball.

If we use the relationship of momentum and momentum, assuming that the time for the two cases is the same and that both start from rest

Bowling ball

           I = F t = Δp

           I = M (v₁ - v₀)

Ping pong ball

           I = F t = Δp

           I = m (v₂ -v₀)

the impulse itself

          M v₁ = m v₂

          v₂ = \frac{M }{ m} v₁

so we conclude that the speed of the ping pong ball is much greater than the speed of the bowling ball.

In conclusion the speed and aceleration of the ping pong ball is greater than that of the bowling ball.

4 0
2 years ago
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