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TEA [102]
3 years ago
11

The state highway patrol radar guns use a frequency of 9.50 GHz. If you're approaching a speed trap driving 37.9 m/s, what frequ

ency shift will your FuzzFoiler 2000 radar detector see
Physics
1 answer:
Grace [21]3 years ago
6 0

The frequency shift your FuzzFoiler 2000 radar detector will see is 10.6 GHz.

<h3>Doppler effect</h3>

The doppler effect show the frequency shift of a source of sound relative to an observer depending on their relative motion.

The frequency shift f' = f(v ± v')/(v ± v") where

  • f = frequency of source = 9.50 GHz,
  • v = speed of sound = 330 m/s,
  • v' = speed of detector = speed of car = 37.9 m/s and
  • v" = speed of source = 0 m/s (since the highway patrol car is stationary)

Since the detector (our car) approaches the source (the radar gun), we use the positive sign in the numerator and a negative in the denominator

So, f' = f(v + v')/(v - v")

Since v" = 0, we have

f' = f(v + v')/(v - 0)

<h3>The frequency shift</h3>

f' = f(v + v')/v

Substituting the values of the variables into the equation, we have

f' = 9.50 GHz(330 m/s + 37.9 m/s)/330 m/s

f' = 9.50 GHz(367.9 m/s)/330 m/s

f' = 9.50 Ghz × 1.115

f' = 10.6 GHz

So, the frequency shift your FuzzFoiler 2000 radar detector will see is 10.6 GHz.

Learn more about Doppler effect here:

brainly.com/question/11419686

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4 years ago
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
3 years ago
What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

6 0
3 years ago
Which is not a step of the scientific method?
Dahasolnce [82]
The six steps of the scientific are:
1. State the question
2. Conduct research
3. Create a hypothesis
4. Perform the experiment
5. Analyze the data
6. Conclusion

So D. would be the correct answer, even though communicating the results could possibly be a step if it's required.
4 0
3 years ago
Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u
tino4ka555 [31]

Answer:

a)  force between them is attraction,   b)  F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

          F =k \frac{q_1q_2}{r^2}

           

In the exercise they give us the values ​​of the loads

          q1 = - 10 mC = -10 10⁻³ C

          q2 = 5 mC = 5 10⁻³ C

           d = 20 cm = 0.20 m

let's calculate

          F = 9 10⁹ \frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}

          F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0
3 years ago
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