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3 years ago

Which phenomenon occurs when a wave encounters a non–transmitting barrier

Physics

2 answers:

SpyIntel [72]3 years ago

6 0

Answer:

Absorption

Explanation:

A non-transmitting barrier would not allow a wave to go through. When a wave is unable to pass through a barrier, it is not transmitted and can get absorbed or reflected back. The wave can also try to go round the barrier.

Most likely, the wave gets absorbed by the barrier and it stops it.

kozerog [31]3 years ago

3 0

Answer:

C. reflection of the wave with the same wave speed

Explanation:

A. refraction of the wave with a different wave speed

B. reflection of the wave with a different wavelength

C. reflection of the wave with the same wave speed

D. refraction of the wave with a different wavelength

Plato said C was right

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3 years ago

The model of the atom proposed by Greek philosophers appears similar to the model proposed centuries later by Dalton. What was t

Sladkaya [172]

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3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

2 years ago

Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.

kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

v² = v₀² - 2 g (y -y₀)

v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

v² = 2g y₀

v = Ra (2g i)

let's calculate

v =√ (2 9.8 16)

v = 17.71 m / s

8 0

3 years ago