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stiv31 [10]
2 years ago
15

What is the radius of the event horizon for a black hole with a mass 7.5 times the mass of the sun? This distance is called the

Schwarzschild radius. Express your answer with the appropriate units.
Physics
1 answer:
Illusion [34]2 years ago
4 0

Answer:

1.1\times10^{4}m

Explanation:

The Schwarzschild radius can be calculated as follows:

R=\frac{GM}{c^2}

Where, <em>G</em> is the gravitational constant, <em>M</em> is the mass and <em>c</em> is the speed of light.

Mass of sun is 1.98×10³⁰ kg

R=\frac{G\times 7.5M_{sun}}{c^2}\\ R=\frac{6.67\times10^{-11}\times(7.5\times1.98\times10^{30})}{(3\times10^8)^2}\\ \Rightarrow R=1.1\times10^{4}m

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Which is a property of light a: mass b: density c: wavelength d: pitch
Reptile [31]
Im pretty sure its C- Wavelength
6 0
3 years ago
Why is it important foe scientists to know how many valence electrons a given element has
Vikentia [17]
Because the number of valence electrons of an element determines the properties and in particular the reactivity of that element.

In fact, elements of the first group (i.e. only one valence electron) have high reactivity, because they can easily give away their valence electron to atoms of other elements forming bonds. On the contrary, elements of the 8th group (noble gases) have their outermost shell completely filled with electrons, so they do not have valence electrons, and they have little or no reactivity at all.
3 0
3 years ago
the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe
Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

          β = 10 log \frac{I}{I_{o}}

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

            I / I₀ = 10 (β / 10)

            I = I₀ 10 (β / 10)

trumpet

            I1 = 1 10⁻¹² 10 (94/10)

            I1 = 2.51 10⁻³ / cm²

Thrombus

           I2 = 1 10⁻¹² 10 (107/10)

           I2 = 5.01 10-2 W / cm²

low

           I3 =1 1-12    (113/10) W/cm²

            I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

           I_total = I₁ + I₂ + I₃

           I_ttoal = 2.51 10-3 + 5.01 10-2 + ​​1.995 10-1

           I_total = 0.00251 + 0.0501 + 0.1995

           I_total = 0.25211 W / cm²

let's bring this amount to the SI system

         β = 10 log (0.25211 / 1 10⁻¹²)

           β = 114 db

7 0
2 years ago
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
What is a neutron start?
nataly862011 [7]

neutron star is formed when the core of a star, that was one ginormous, collapse on its own mass thus it loses it's volume exponentially (y=a^x) while the mass of the star decreases linearly (y=mx+c) thus the neutron star has mass of 10 to 29 sun in the radius of 10 km( for scale earth has 6731 km) thus it's density is super high!!!!!!!!!!!!!!!!!!!

hope it helps you! ;)

7 0
3 years ago
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