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sergiy2304 [10]
3 years ago
5

You drive your car at a speed 40km/h for 1 hour, then slow down to 30km/h for the next 20km. How far did you drive, and what was

your average speed?
Physics
1 answer:
Bingel [31]3 years ago
8 0
Speed=distance/time, rearranging gives speed*time=distance therefore the first section of the journey yielded a total distance of 40km, the second journey was stated to be 20km which was completed in (20/30)*60=40minutes. Therefore the total journey was 60km which was completed in 1hr40 minutes so the average speed was (60/1.66..)=36km/h.
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When I went through with the math, the answer I came upon was: 
<span>6.67 X 10^14 </span>

<span>Here is how I did it: First of all we need to know the equation. </span>

<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
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<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>

<span>Divide the speed of light by the wavelength. </span>
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<span>Answer: 6.67 X 10^14 s- hope this helps</span>
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3 years ago
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ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

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v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

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