By what factor must you increase the intensity of a sound in order to hear a 1.0-dB rise in the sound level?

Physics

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

The right approach will be "1.3". A further explanation is given below.

Explanation:

As we know,

In Decibels, the change in sound volume will be:

= $10log\frac{I_1}{I_2}$

Now,

According to the question,

⇒ $1=10 log \frac{I_1}{I_2}$

By applying cross multiplication and putting the value of log, we get

⇒ $10^{\frac{1}{10} }=\frac{I_1}{I_2}$

⇒ $1.26=\frac{I_1}{I_2}$

⇒ $\frac{I_1}{I_2}=1.3$

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A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns

Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

$f'=\frac{V+V_{o}}{V-V_{s}} f$ (1)

Where:

$f=250,000 Hz$ is the actual frequency of the sound wave

$f'=240,000 Hz$ is the "observed" frequency

$V=340 m/s$ is the velocity of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which is the balloon

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$ (2)

$V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}$ (3)

$V_{s}=-14.16 m/s$ (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ( $t=4.8 s$ ) given as data, we can find the distance: