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SVEN [57.7K]
3 years ago
9

Which statements accurately describe the Doppler effect?

Physics
2 answers:
Mnenie [13.5K]3 years ago
8 0

Answer:

A. It is used to learn about the behavior of the universe.

B. It was used by Hubble to measure velocities of galaxies.

C. It indicates galaxy motion through light wavelengths.

chubhunter [2.5K]3 years ago
5 0

The doppler effect is the increase or decrease in the frequency of sound, light, or other waves as the source and observer move toward or away from each other.

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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0
3 years ago
Which descriptions best fit the labels? X: kinetic energy Y: potential energy X: potential energy Y: kinetic energy X: mechanica
GalinKa [24]

its the 3rd option!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

4 0
3 years ago
Read 2 more answers
How do unicellular organisms grow?
Snezhnost [94]

Answer:

C is your answer good luck

Explanation:

3 0
3 years ago
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Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The
Alexxx [7]

Answer:

The power of top half of the lens is 0.55 Diopters.

Explanation:

Since, the person can see an object at a distance between 34 cm and 180 cm away from his eyes. Therefore, 180 cm must be the focal length of the upper part of lens, as the top half of the lens is used to see the distant objects.

The general formula for power of a lens is:

Power = 1/Focal Length in meters

Therefore, for the top half of the lens:

Power = 1/1.8 m

<u>Power = 0.55 Diopters</u>

3 0
3 years ago
Physicists often use a different unit of energy, the electron volt, when dealing with energies at the atomic level. One electron
Alex Ar [27]

Answer:

a)  E = 1.06 10⁻¹⁹ J,     b)   v = 9.78 10⁵ m / s

Explanation:

The physical magnitudes can be given in several units, but in general all must be reduced to the same system in a given exercise, the most used system is the international (SI)

           1 eV =q V=  1.6 10⁻¹⁹ J

let's reduce the quantities requested

a) E = 1.0 eV to Joule

           E = 1.0 eV (1.6 10⁻¹⁹ J / 1 eV)

           E = 1.06 10⁻¹⁹ J

b) the kinetic energy is given by

           K = ½ m v²

           v = \sqrt{\frac{2K}{m} }

the mass of the proton is

          m = 1,673 10⁻²⁷ kg

let's reduce the energy to the SI system

           E = 5000 ev (1.6 10-19 J / 1 eV) = 8000 10⁻¹⁹ J

let's calculate

           v = \sqrt{ \frac{2 \ 8000 \ 10^{-19}}{1.673 \ 10^{-27} }  }

           v = \sqrt{ 95.637 \  10^{10} }

           v = 9.78 10⁵ m / s

3 0
2 years ago
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