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e-lub [12.9K]
3 years ago
15

Identify the techniques used to try to locate extra solar (exo) planets

Physics
2 answers:
attashe74 [19]3 years ago
8 0

Answer:

the doppler technique

Explanation:

The doppler technique was  an indirect method for finding extrasolar planets and brown dwarfs from radial-velocity measurements using observation of Doppler shifts in the spectrum of the planet's parent star.

antiseptic1488 [7]3 years ago
5 0

Answer:

A Transits

B Doppler technique

Explanation:

The techniques used for the location of extra solar planets are both transit method and the Doppler technique. Extra solar planets are those that orbit a star other than the Sun of our solar system.

Transit method uses the method of measuring light curve from the stars and the reduction of the brightness over time is noted. This is an indirect method.

Doppler technique is a much more  reliable method  to discover exoplanets. It is also called wobble method. This is much more reliable to detect exoplanets. When there is no exoplanet for a given star, the Doppler shift pattern does not change. If exoplanet exists, it changes.

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A rod (length = 2.0 m) is uniformly charged and has a total charge of 5.0 nC. What is the electric potential (relative to zero a
Ksenya-84 [330]

Answer:

The electric potential is  V  =  15.6 V

Explanation:

From the question we are told that

   The length  of the rod is  L =  2.0 \ m

     The total charge of the rod is  q =5.0 nC = 5.0*10^{-9} C

      The length from the center is  d =  3.0 \ m

The diagram illustrating the setup for this question is shown on the first uploaded image

From the diagram the potential at point  A  dl is mathematically represented as

         dV  =  K  \frac{dq}{l}

Where K is the coulomb constant with a value  K  = 9*10^9 \   Nm^2 /C^2

where q is the charge in charge  the rod relative to its distance from A  is mathematically represented as

         dq =  \frac{q}{L}  dl

This a small unit length of the rod

So         V = \frac{q}{L}  \int\limits^4_2  {\frac{dl}{l} } \,

        =>   V =  k\frac{q}{L}  ln [\frac{4}{2} ]

              V =  k\frac{q}{L}  ln2

Substituting values

                V  =  9* 10^9  *  \frac{5*10^{-9}}{2} * ln 2

                 V  =  15.6 V

         

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4 years ago
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Digiron [165]

<u>Answer;</u>

<em>D.  The object’s weight changes, but its mass stays the same.</em>

<u>Explanation;</u>

  • Mass is the amount of matter in a object, which is measured in kilograms. Mass of an object is measured using a beam balance. It is important to note that the mass of an object or a body remains constant, and does not vary from one place to another. For instance the mass of a person on the moon will be the same as when the person is on the earth surface.
  • Weight on the other hand is the measurement of gravitational pull of an object. weight is measured using a spring balance and measured in Newtons. Weight varies from one place to another depending on the gravitational pull of a given surface.
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anzhelika [568]

Answer:

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Explanation:

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Think - more energy, bigger waves, more waves, and higher sound

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<u>B</u> option is correct    .

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