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il63 [147K]
3 years ago
13

If you detected radio signals with an average wavelength of 67 cm and suspected that they came from a civilization on a distant

Earth-like exoplanet, roughly how much of a change in wavelength (in cm) should you expect to detect as a result of the orbital motion of the distant exoplanet?
Physics
1 answer:
Crazy boy [7]3 years ago
8 0

Answer:

0.0066553 cm

Explanation:

v = Earth's average orbital velocity = 29.8 km/s

c = Speed of light = 3\times 10^8\ m/s

\lambda = Wavelength = 67 cm

\Delta\lambda = Change in wavelength

The Doppler shift is given by

\dfrac{v}{c}=\dfrac{\Delta \lambda}{\lambda_0}\\\Rightarrow \Delta \lambda=\dfrac{v\lambda_0}{c}\\\Rightarrow \Delta \lambda=\dfrac{29800\times 0.67}{3\times 10^8}\\\Rightarrow \Delta\lambda=0.000066553\ m=0.0066553\ cm

The change in wavelength is 0.0066553 cm

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Alexxandr [17]

C.joined forces

your answer

7 0
3 years ago
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In physical science lab, Ben and Jerry added small pieces of magnesium to hydrochloric acid. They noticed that bubbles formed, t
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4 years ago
a satellite maintains an orbit equidistant from the earth at all points along its orbital path hiw is the satellite affected by
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5 0
4 years ago
A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w
garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

E_s=\dfrac{1}{2}kx^2

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

E_P=mgx

Equating these two energies,

\dfrac{1}{2}kx^2=mgx

\dfrac{k}{m}=\dfrac{2g}{x}.............(1)

The expression for the oscillation frequency is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}} (from equation (1))

f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0
4 years ago
2. What is the momentum of a football of mass 500g traveling at a
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Answer:

5000 ns

Explanation:

hope this helps

5 0
3 years ago
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