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3 years ago

8

When frequency is increased what happens to the amplitude?

1 answer:

Paul [167]3 years ago

6 0

Answer: For sound, a higher frequency corresponds to a higher pitch while a lower frequency corresponds to a lower pitch. Amplitude corresponds to the loudness of the sound. Since sound at all frequencies has the same speed in air, a change in frequency means a change in wavelength.

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How are eras and periods of the geologic time scale named?

Natalka [10]

Answer:

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Explanation:

In the early 1800's a system for naming geologic time periods was devised using four periods of geologic time. They were named using Latin root words, Primary, Secondary, Tertiary and Quaternary. ... Keep in mind that this chart is focused on geologic time periods. There are also geologic Eons, Eras, and epochs.

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3 years ago

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Distance = 500 miles<br><br> Time = 10 hours<br><br> Average Speed = ?????

kotegsom [21]

Average speed= 50mph

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3 years ago

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Answer:

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3 years ago

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A person has a gravitational force (weight) of 600 N on Earth. Suppose the mass of the Earth is double and the radius shrinks to

Alenkasestr [34]

Explanation:

that's impossible,the radius of the earth can't decrease when the mass doubles!

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3 years ago

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

$\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$ ......(1)

Here $\lambda_{s}$ is source wavelength, $\lambda_{o}$ is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, $\lambda_{s}$ = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, $\lambda_{0}$ = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

$555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$

$\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84$

${\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056$

$1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}$

$\frac{v}{c}=\frac{0.2944}{8.056}$

$v = 0.036c=0.036\times3\times10^{8}$

v = 1.08 x 10⁷ m/s

8 0

3 years ago