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Mekhanik [1.2K]
3 years ago
9

A highway patrol officer uses a device that measures the speed of vehicles by bouncing radar off them and measuring the Doppler

shift. The outgoing radar has a frequency of 100 GHz and the returning echo has a frequency 15.0 kHz higher. What is the velocity of the vehicle? Note that there are two Doppler shifts in echoes. Be certain not to round off until the end of the problem, because the effect is small.
Physics
1 answer:
Whitepunk [10]3 years ago
7 0

Answer:

V = 2.5725*10^{-5}m/s

Explanation:

The data we are given is:

V = ?     fo = 100GHz        fs = 100000015KHz    C = 343m/s

With the doppler effect formula we can calculate the frequency perceived by the vehicle as:

fr = \frac{C+V}{C}*fo     (1)

Then, we sound waves bounce back on the vehicle:

fs = \frac{C}{C-V}*fr      (2)   Replacing tha value obtained in (1)

fs = \frac{C}{C-V}*\frac{C+V}{C}* fo   Solving for V:

V = C*\frac{fs-fo}{fs+fo}=2.5725*10^{-5}m/s

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Hope this helps
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Does the atmosphere contain 21% nitrogen in 78% oxygen
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7 0
3 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
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