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Papessa [141]
3 years ago
9

If the sun were to collapse into a black hole, what would the radius be?

Physics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

It would be 1/3rd size of a neutron star

Explanation:

the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass.

Marianna [84]3 years ago
4 0

Answer:

If the Sun, with its mass of 1 MSun, were to become a black hole the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass

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What relationship exists between the amplitude of a wave and amount of disturbance in the water?
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A wave is a result of the disturbance in the equilibrium state.  There are two types of wave, transverse and longitudinal. Transverse wave affects amplitude while longitudinal wave affects the frequency of the wave. As for the transverse wave, the magnitude of the perpendicular disturbance of the wave is directly proportional to the amplitude of the wave. The higher the transverse disturbance the higher the amplitude.
6 0
3 years ago
Where is the potential energy equal to zero?
s2008m [1.1K]

Answer:

im sure your already past this but it's E.

Explanation:

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4 0
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Which of the following is NOT a potential result of climate change?
stepladder [879]

Answer:

My answer :

Explanation:

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4 0
3 years ago
What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight ou
Lubov Fominskaja [6]

Answer:

The torque about his shoulder is 34.3Nm.

The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.

Explanation:

The solution to the problem can be found in the attachment below.

6 0
3 years ago
A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a
andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

sin\theta = 1.22\frac{\lambda}{d}

Here,

\lambda = Wavelength

d= Diameter of aperture

\theta = Angular resolution or diffraction angle

Our values are given as,

\theta = 11\°

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The speed of the sound is v = 343 m/s

The wavelength of the sound is

\lambda = \frac{v}{f}

Here,

v = Velocity of the wave

f = Frequency

Replacing,

\lambda = \frac{(343 m/s)}{(9100 Hz)}

\lambda = 0.0377 m

The diffraction condition is then,

sin\theta = 1.22\frac{\lambda}{d}

Replacing,

sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}

d = 0.24 m

Therefore the diameter should be 0.24m

6 0
3 years ago
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