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2 years ago

When a wave is too steep to support itself, the wave front collapses creating a ________ that advances up the shoreline.

Physics

1 answer:

galben [10]2 years ago

8 0

When a wave is too steep to support itself, the wave front collapses therefore creating a break.

This is defined as the propagation of disturbance from one place to another in an organized manner.

In situations where the wave is too steep to support itself there is a break in the wavefront which advances up the shoreline.

Read more about Wave here brainly.com/question/6069116

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When the Moon orbits Earth, what is the centripetal force?

nata0808 [166]

Answer:

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2 years ago

Oscilloscope amplitude and frequency problem. Study the above graph. The volts/div dial is set to 2 volts/div and the time/div d

denis-greek [22]

Answer:

Amplitude = 8 Volts

Frequency = 0.067 kHz

Explanation:

Note: The missing picture in question is attached for your review.

Given:

Volts/Div = 2 V/div

Time/Div = 5 msec/div

Finding Amplitude:

Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

$Amplitude = 4 div/volts * 2 volts/ div )\\Amplitude = 8 Volts$

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)

Finding Frequency:

As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

$Time Period = 3 div * 5msec /div\\Time Perod = 15 msec$

Since,

$Frequency = \frac{1}{Time Period}\\Frequency = \frac{1}{15m}\\Frequency = 0.067 kHz$

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3 years ago

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3 years ago

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The c-cl bond dissociation energy in cf3cl is 339 kj/mol. What is the maximum wavelength of photons that can rupture this bond?

ruslelena [56]

Answer:

3.53*10^{-7} m

Explanation:

Photon that can rupture the bonds are those with the energy of the bond dissociation energy. If we want to know the energy for each molecule we have to take into account that:

$1mol=6.022*10^{23}molecule$

Hence, we have

$E_d=339\frac{10^{3}J}{mol}*\frac{1mol}{6-022*10^{23}molecules}=5.62*10^{-19}J/molecule$

but the energy is also:

$E_d=h\nu =\frac{hc}{\lambda}\\\\\lambda=\frac{hc}{E_d}$

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$\lambda=\frac{(6.62*10^{-34}Js)(3*10^{8}m/s)}{5.62*10^{-19}J}=3.53*10^{-7}m$

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