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Nimfa-mama [501]
3 years ago
7

What is the low end of the range of surface temperature of white stars

Physics
1 answer:
Lelechka [254]3 years ago
8 0
The way astronomers determine the surface temperature<span> of a </span>star<span> which is determined by the spectral lines present in a </span>stars<span> spectrum</span>
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A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio
docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

Explanation:

- write the equation F(r) = -Kr^{4} with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

3 0
3 years ago
CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY!
BARSIC [14]
<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />

According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.  

Now, in this case we are on Earth, so the gravity value is 9.8\frac{m}{s^{2}}  

Note the objects experience the acceleration of gravity regardless of their mass.

Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely  true on Earth, unless the following condition is fulfiled:

If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at 9.8\frac{m}{s^{2}} and hit the ground at approximately the same time.  


5 0
3 years ago
At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n
Andreyy89
Below are the choices that can be found elsewhere:

 a. 268 kJ 
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have: 

<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>

<span>Ga (l) --> Ga (g) delta H = 271 kJ/mol </span>

<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>

<span>Ga (s) --> Ga (l) delta H = 6 kJ/mol </span>

<span>At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). </span>

<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>
7 0
3 years ago
A 40W lamp wastes 34 J of energy every second by heating its surroundings.
Artemon [7]

Answer:

15\%.

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

4 0
2 years ago
What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?In meters
Leto [7]
Apply:
wavelength = speed/frequency
= 350 m/s : 140 Hz = 2.5 m.
3 0
3 years ago
Read 2 more answers
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