14. A tuning fork vibrates at a frequency of

Physics

1 answer:

cupoosta [38]3 years ago

4 0

Answer:

Every time you strike a tuning fork, you're setting off a tiny, invisible hurricane. Thrashing back and forth at tremendous speeds, the two prongs of the fork, known as "tines," are smashing against nearby air molecules, kicking off a chain of impacts that echo through the air. When these violent, microscopic collisions hit your eardrum, your brain processes them as a gentle hum.

By hitting a tuning fork, you're causing its tines to vibrate back and forth several hundred times per second. Often, the vibrations are so fast that they're not visible to the human eye. If you need proof, simply dip a humming tuning fork into a cup of water -- it'll kick up a surprisingly large jet of water. In scientific terms, the speed of a tuning fork's vibrations is known as its frequency, a quantity measured in hertz (Hz), or vibrations per second.

The way a tuning fork's vibrations interact with the surrounding air is what causes sound to form. When a tuning fork's tines are moving away from one another, it pushes surrounding air molecules together, forming small, high-pressure areas known as compressions. When the tines snap back toward each other, they suck surrounding air molecules apart, forming small, low-pressure areas known as rarefactions. The result is a steady collection of rarefactions and compressions that, together, form a sound wave.

The faster a tuning fork's frequency, the higher the pitch of the note it plays. For instance, for a tuning fork to mimic the top key on a piano, it needs to vibrate at 4,000 Hz. To mimic the lowest key, on the other hand, it would only need to vibrate at 28 Hz.

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a) E = 0

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We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

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q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

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However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

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