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2 years ago

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If Two stars, Betty and Wilma, are both on the main sequence. Betty is more luminous than Wilma. Which one has a hotter surface

temperature? Which one is more massive? Which one is bigger? If they both formed at the same time, which one will evolve off the main sequence first?

1 answer:

monitta2 years ago

5 0

Answer:

Betty

Explanation:

Since they're both on the most sequence but Betty is more luminous than Wilma, Betty must be located to a higher place on the most sequence. Therefore, Betty encompasses a hotter surface temperature, is more massive, and encompasses a larger radius. Betty also will evolve faster than Wilma, and if they were formed at the identical time Betty will put off the most sequence first

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A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

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Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, $\Delta f = 4\ beats/s$

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

$\Delta f = f - f_{i}$

$f_{i} = f \pm \Delta f$

when

$f_{i} = f + \Delta f = 523 + 4 = 527 Hz$

when

$f_{i} = f - \Delta f = 523 - 4 = 519 Hz$

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

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24 How long does it take for the Earth to orbit the Sun?

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Answer:

365 days

So The Final Answer is a Year

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Explanation:

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2 years ago

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A star is measured to be 18.9 light years from earth. what is the correct answer in the calculation shown below? (consider both

Lapatulllka [165]

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Therefore, 18.9 light years is equal to 1.79 x 10^17 meters.

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2 years ago

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A 75 kg object is moving west at 5.6 m/s. What is the object's momentum? (p = mv)

Licemer1 [7]

Explanation:

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<em>hope</em><em> it</em><em> was</em><em> helpful</em><em> to</em><em> you</em>

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2 years ago

A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.

ruslelena [56]

Answer:

C. $W = 115.12\,Btu$

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process ( $\eta$ ), no unit, is:

$\eta = \left(1-\frac{T_{L}}{T_{H}} \right)$ (1)

Where:

$T_{L}$ - Temperature of the cold reservoir, measured in Rankine.

$T_{H}$ - Temperature of the hot reservoir, measured in Rankine.

If we know that $T_{H} = 1809.67\,R$ and $T_{L} = 584.67\,R$ , then the energy efficiency of the ideal thermal process is:

$\eta = 0.678$

By First Law of Thermodynamics, we calculate the work output:

$W = Q_{H}-Q_{L}$

$W = \frac{W}{\eta} -Q_{L}$ (By definition of efficiency)

$Q_{L} = \frac{W}{\eta}-W$

$Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W$ (2)

Where:

$Q_{H}$ - Heat received by the engine, measured in Btu.

$Q_{L}$ - Heat rejected by the engine, measured in Btu.

$W$ - Work output, measured in Btu.

If we know that $\eta = 0.678$ and $Q_{L} = 55\,Btu$ , then the work output of the Carnot engine is:

$W = \frac{Q_{L}}{\frac{1}{\eta}-1 }$

$W = 115.807\,Btu$

The work output of the Carnot engine is 115.807 Btu. (Answer: C)

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2 years ago