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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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EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} where;

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If the magnification is - 0.6

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v = -0.5u

since v = 10cm

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Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm

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7 0
3 years ago
In a lab, a student drags a shoe across the floor at constant speed. If the coefficient of static friction between the floor and
Svetllana [295]
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3 years ago
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antiseptic1488 [7]
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3 years ago
A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th
Reika [66]

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

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In the formula, A and B are constant, so the behaviour depends only on the function cos \theta. The value of cos \theta are:

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- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B


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3 years ago
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