Is the water the same temperature in alaska and mexico?

If the bat emits a sound at a frequency of 80.2 kHzkHz and hears it reflected at a frequency of 83.5 kHzkHz while traveling at a

solong [7]

To solve this problem we will apply the concepts related to the Doppler Effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other.

Mathematically it can be described as

$v' = v (\frac{f' ( v - v_{bat}) - f ( v + v_{bat})}{f'(v-v_{bat}) + f ( v + v_{bat})})$

Where,

v = Speed of sound

f' = Reflected frequency

f = Emitted Frequency

$v_{bat}$ = Speed of the object

Assuming that the speed of the medium is the speed of sound of 343m / s, we will then have that:

$v' = 343 (\frac{(83.5) (343 -3.9) - 80.2(343+3.9)}{83.5(343-3.9) + 80.2(343+4.3)})$

$v' = 3.0134m/s$

8 0

3 years ago

lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of

bonufazy [111]

Answer:

The intensity of laser 2 is 4 times of the intensity of laser 1.

Explanation:

The intensity in terms of electric field is given by :

$U=\dfrac{1}{2}\epsilon_o E^2$

E is electric field

It means, $U\propto E^2$

In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.

Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

$\dfrac{U}{U'}=(\dfrac{E}{E'})^2$