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jok3333 [9.3K]
3 years ago
14

A) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies

, even if the stellar radius and distance are unknown.
b) Explain why in practice this method does not work well if both frequencies are on theRayleigh-Jeans side of the spectrum,hνkT.
c) Derive a simple, approximate expression for the temperature when both measurements areon the Wien tail,hνkT.
d) Derive an expression for the star’s radius if a distance measurement is also available (e.g.,from parallax).

Physics
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

The answers to the questions have been solved in the attachment.

Explanation:

The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.

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photoshop1234 [79]

The answer is 0.245N.

<h3>What is kinetic energy?</h3>
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(b) 0.100

For the block on the left, f_{k} =u_{k} n= 0.100(2.45N)=0.245N.

∑F_{x}=ma_{x}

–0.308N+0.245N=(0.250kg)a

a=−0.252m/s^{2} if the force of static friction is not too large.

For the block on the right, f_{k} =u_{k} n=0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

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Two sets of Christmas lights are available. For set A, when one bulb is removed, the remaining bulbs remain illuminated. For set
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In Set A, the bulbs (resistors) are connected  parallely to each other, this means that even if one of the bulbs fuses or removed, the circuit will still be completed and others continue to glow.

And in parallel connection if the resistance of the two resistors are same  powered delivered to each is same.

In Set B, bulbs are in series connection, this means that when one of the bulb is removed or fuses, the circuit will break and other bulbs can not operate.In this situation as well  if the resistance of two resistors is same then the power delivered is same.

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Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}

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Answer:

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