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yanalaym [24]
3 years ago
14

Tectonic plates are large segments of the Earth's crust that move slowly. Suppose that one such plate has an average speed of 4.

0 cm/year. (a) What distance does it move in 1 s at this speed? (b) What is its speed in kilometers per million years?
Physics
1 answer:
Kay [80]3 years ago
3 0

Answer:

a) 1.268×10⁻⁷ cm/sec

b) 40 km/million years.

Explanation:

Speed of tectonic plate = 4.0 cm/year

a) Convert the year to seconds

1 year = 365 days × 24 hours × 60 minutes × 60 seconds

⇒1 year = 31536000 seconds

So,

4\ cm/year = \frac{4}{31536000}\\\Rightarrow 4\ cm/year = \frac{1}{7884000}\\\Rightarrow 4\ cm/year = 1.268\times 10^{-7} cm/sec

Distance plate will move in 1 second is 1.268×10⁻⁷ cm/sec

b) Convert cm to km

4\ cm=\frac{4}{10^{-5}}\ km

So, in 1 year it will move 0.00004 km

Hence in 1 million years is will move

0.00004 × 10⁶ = 40 km/million years.

Speed in kilometers per million years is 40 km/million years.

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Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

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But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

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