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Dafna1 [17]
2 years ago
6

A sound wave traveling in water at 144m/s has a wavelength of 0.5m determine the frequency of the wave

Physics
2 answers:
mrs_skeptik [129]2 years ago
4 0

Heya!!

For calculate frequency, lets applicate formula:

                                                       \boxed{f=v/\lambda}

                                                   <u>Δ   Being   Δ</u>

                                               f = Frequency = ?

                                           v = Velocity = 144 m/s

                                          \lambda = Wavelenght = 0,5 m

⇒ Let's replace according the formula:

\boxed{f = 144\ m/s / 0,5\ m }

⇒ Resolving

\boxed{f = 288\ Hz}

Result:

The frequency of that wave is <u>288 Hertz</u>

Good Luck!!

Setler79 [48]2 years ago
3 0

Answer:

sadasdsadasd

Explanation:

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A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

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With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

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Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

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