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Agata [3.3K]
3 years ago
6

While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r

ange of wavelengths of this sound?
Physics
1 answer:
sdas [7]3 years ago
6 0

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m

If f = 78 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

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A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab
maria [59]

Answer:

W = 0.678 rad/s  

Explanation:

Using the conservation of energy:

E_i =E_f

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

\frac{1}{2}IW^2+ \frac{1}{2}mV^2 = mgh

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I =\frac{2}{3}mR^2

where m is the mass and R the radius, so:

I =\frac{2}{3}(0.426kg)(11.3m)^2

I = 36.26 Kg*m^2

Then, replacing values on the initial equation, we get:

\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)V^2 = (0.426kg)(9.8)(5m)

also we know that:

V =WR

so:

\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)W^2R^2 = (0.426kg)(9.8)(5m)

Finally, solving for W, we get:

W^2(\frac{1}{2}(36.26)+ \frac{1}{2}(0.426kg)(11.3m)^2) = (0.426kg)(9.8)(5m)

W = 0.678 rad/s

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3 years ago
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