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makvit [3.9K]
3 years ago
7

Help please number 8 an 9

Physics
1 answer:
Murrr4er [49]3 years ago
5 0
8) The reason is that MRIs use radio-waves, which are electromagnetic waves with lower frequency, and so lower energy, than X-rays.
In fact, typical frequencies for radio waves are f=10-10^9 Hz, while X-rays have typical frequencies of 10^{16} - 10^{19} Hz. Since the energy of a wave is given by
E=hf
where h is the Planck constant and f the frequency, we see that the lower the frequency, the lower the energy. Therefore, radiowaves have less energy than X-rays, so they are less ionising and less dangerous than X-rays.

9) The false statement is:
a) gamma-rays are low-frequency waves.
It's actually the opposite, in fact: gamma rays are the electromagnetic waves with higher frequencies (above 10^{20}Hz).

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What describes how fast an object is moving and in what direction
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A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a
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Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

I=\frac{P}{4\pi r^2}

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0
3 years ago
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