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nalin [4]
3 years ago
9

The Schwarzschild radius is the distance from an object at which the escape velocity is equal to the speed of light. A black hol

e is an object that is smaller than its Schwarzschild radius, so not even light itself can escape a black hole. The Schwarzschild radius ???? depends on the mass ???? of the black hole according to the equation ????=2????????????2 where ???? is the gravitational constant and ???? is the speed of light. Consider a black hole with a mass of 5.7×107M⊙. Use the given equation to find the Schwarzschild radius for this black hole. Schwarzschild radius: m What is this radius in units of the solar radius
Physics
1 answer:
solmaris [256]3 years ago
7 0

1. 1.69\cdot 10^{11} m

The Schwarzschild radius of an object of mass M is given by:

r_s = \frac{2GM}{c^2} (1)

where

G is the gravitational constant

M is the mass of the object

c is the speed of light

The black hole in the problem has a mass of

M=5.7\cdot 10^7 M_s

where

M_s = 2.0\cdot 10^{30} kg is the solar mass. Substituting,

M=(5.7\cdot 10^7)(2\cdot 10^{30}kg)=1.14\cdot 10^{38} kg

and substituting into eq.(1), we find the Schwarzschild radius of this black hole:

r_s = \frac{2(6.67\cdot 10^{-11})(1.14\cdot 10^{38} kg)}{(3\cdot 10^8 m/s)^2}=1.69\cdot 10^{11} m

2) 242.8 solar radii

We are asked to find the radius of the black hole in units of the solar radius.

The solar radius is

r_S = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m

Therefore, the Schwarzschild radius  of the black hole in solar radius units is

r=\frac{1.69\cdot 10^{11} m}{6.96\cdot 10^8 m}=242.8

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Alright, let’s break this down. There are three resistors in this circuit, meaning that we have to find the equivalent resistance. Luckily, they are all in parallel with one another; this means we can add the resistances together without having to do inverses like in a series problem. This means that the equivalent resistance, Req, would equal:

Req=R1 + R2 + R3

Req=6 + 14+ 10

Req=30 Ohms

This means that we could theoretically replace all three resistors with a 30 Ohm resistor and accomplish the same goal. Now, the entire voltage of the system would normally be reduced to zero after passing through the resistors - in this case, the 60 Vs would be lost after passing through 30 Ohms. This means we’re losing 2V/Ohm; now we can figure out how much we’re losing at each resistor.

By losing 2V per Ohm, we’re losing 12 V at the first resistor, 28 V at the second resistor, and 20 V at the third resistor.

Finally, we can calculate the current through the circuit; for a series circuit, the current remains the same. Using V=IR, we can find that:

V=IR

60 V = I(30 Ohms)

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The current passing through the circuit is 2 Amps.

Hope this helps!

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Explanation:

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