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4 years ago

16)

Physics

1 answer:

VMariaS [17]4 years ago

6 0

Answer:

a.proton, proton

hope this helped :) have a goodday

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The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

BigorU [14]

1. Frequency: $7.06\cdot 10^{14} Hz$

The frequency of a light wave is given by:

$f=\frac{c}{\lambda}$

where

$c=3\cdot 10^{-8} m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

In this problem, we have light with wavelength

$\lambda=425 nm=425\cdot 10^{-9} m$

Substituting into the equation, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz$

2. Period: $1.42 \cdot 10^{-15}s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

The frequency of this light wave is $7.06\cdot 10^{14} Hz$ (found in the previous exercise), so the period is:

$T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s$

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Mental processes refers to

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Internal,covert processes

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What wave moves in the same direction as the force that created it?

SIZIF [17.4K]

Longitudinal waves have movement in the direction of the wave energy.

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The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

Elanso [62]

On Earth, the period of a pendulum is given by:
$T_{earth}=2\pi \sqrt{ \frac{L}{g_{earth} }$
where L is the length of the pendulum and $g_{earth}=9.81~m/s^2$ is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
$T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }$
where $g_{mars}=3.71~m/s^2$ is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
$\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} } = \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63$
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.

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A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magneti

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Answer:

Explanation: please see attached file I attached the answer to your question.

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