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3 years ago

If a wave has amplitude of 2 meters, a wavelength of 5 meters, and a frequency of 60 Hz, at what speed is it moving?

Physics

2 answers:

kow [346]3 years ago

6 0

Answer: The velocity is 300 meters per second.

Explanation: The velocity of a wave is related to the frequency and the wavelenght by the following equation:

velocity = frequency*wavelenght

or, using the formal notation;

v = f*λ

We know that f = 60Hz (and Hz is 1/s) and λ = 5m, so we have:

v = 5m*60(1/s) = 300m/s

klemol [59]3 years ago

5 0

The relationship between speed of a wave w, frequency f and wavelength $\lambda$ is
$v = \lambda f$
For the wave of our problem, $\lambda = 5 m$ and f=60 Hz, so its speed is
$v=(5 m)(60 Hz)=300 m/s$

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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

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