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castortr0y [4]
3 years ago
7

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the

Earth. With such a small diameter it must be a compact object, and with such a large mass it can't be a white dwarf or a neutron star, so a black hole is the only possibility remaining. How was the diameter of Cygnus X-1 found
Physics
1 answer:
SOVA2 [1]3 years ago
7 0

Answer:

Answer is It was deduced from the rate at which it glimmers.

Refer below.

Explanation:

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the Earth. With such a small diameter it must be a compact object, and with such a large mass it can't be a white dwarf or a neutron star, so a black hole is the only possibility remaining. The diameter of Cygnus X-1 found:

It was deduced from the rate at which it glimmers.

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A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.
fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe,  the velocity is given by

v=\frac {2Lf}{n}

Making L the subject then

L=\frac {nV}{2f}

Where f is the frequency,  L is the length,  n is harmonic number,  v is velocity

Substituting 1 for n,  320 Hz for f and 331 m/s for v then

L=\frac {1*331}{2*320}=0.5171875\approx 0.52m

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, f2=2\times \frac {v}{2L} and f3=3\times \frac {v}{2L}

f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz

(c)

When v=367 m/s then

f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz

5 0
3 years ago
A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling
KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

                    = 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0
3 years ago
Help, please
Mice21 [21]

Answer:

A and A

Explanation:

see paper for work! :)

8 0
3 years ago
A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charl
Bond [772]

Answer:

Boyle's Law

\therefore P_1.V_1=P_2.V_2

Explanation:

Given that:

<u><em>initially:</em></u>

pressure of gas, = P_1

volume of gas, = V_1

<em><u>finally:</u></em>

pressure of gas, = P_2

volume of gas, = V_2

<u>To solve for final volume</u> V_2

<em>According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.</em>

<em>According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.</em>

But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.

Here nothing is said about the temperature, so we consider the Boyle's Law which states that <em>at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.</em>

Mathematically:

P_1\propto \frac{1}{V_1}

\Rightarrow P_1.V_1=k\ \rm(constant)

\therefore P_1.V_1=P_2.V_2

5 0
4 years ago
Example of items that changed chemically
Ostrovityanka [42]
Burning of gases is one the example of chemical change
3 0
3 years ago
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