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3 years ago

As demonstrated by the Doppler Effect, why does sound increase in pitch as a sound source approaches you?

Physics

2 answers:

mixas84 [53]3 years ago

6 0

Answer: D

Explanation: I took the test

scZoUnD [109]3 years ago

3 0

I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.

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Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When t

stealth61 [152]

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

$v_{oA} = 20 m/s$ : Initial speed of cyclist A

$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

$t_{A} =t_{B} = 12s$

time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t\\$

$36= 12 + a_{B} *12$

$a_{B} = \frac{36-12}{12}$

$a_{B} = 2 \frac{m}{s^{2} }$

$d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}$

$d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}$

$d_{B} = 288m$

Cyclist A Kinematics

$d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}$

$d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}$

$d_{A} = 240 + 72*(a_{A})$

$288=240+72*(a_{A} )$

$a_{A} = \frac{288-240}{72}$

$a_{A} = 0.67 \frac{m}{s^{2} }$

$v_{fA} = v_{oA} + a_{A} * t$

$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$

7 0

3 years ago

An optical fiber made of glass with an index of refraction 1.50 is coated with a plastic with index of refraction 1.30. What is

Helen [10]

To solve this problem we will use Snell's law. This law is used to calculate the angle of refraction of light by crossing the separation surface between two means of propagation of light (or any electromagnetic wave) with a different index of refraction.

$n_1 sin\theta_1 = n_2 sin\theta_2$

Where,

$n_{1,2}$ = Index of refraction for each material

$\theta_1$ = Angle of incidence and refraction

A ray of light propagating in a medium with index of refraction $n_1$ hitting an angle $\theta_1$ on the surface of a medium of index $n_2$ with $n_1> n_2$ can be fully reflected inside the medium of highest refractive index: