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Dmitry [639]
3 years ago
5

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

ar. Express your answer using two significant figures.2.) A 6 MSun black hole that formed in the supernova of a massive star. Express your answer using two significant figures.3.) A mini-black hole with the mass of the Moon. Express your answer using two significant figures.4.) Estimate the Schwarzschild radius (in kilometers) for a mini-black hole formed when a superadvanced civilization decides to punish you (unfairly) by squeezing you until you become so small that you disappear inside your own event horizon. (Assume that your weight is 50 kg.) Express your answer using one significant figure.
Physics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

(I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

Explanation:

Given that,

Mass of black hole m= 1\times10^{8} M_{sun}

(I). We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=2.94\times10^{8}\ km

(II). Mass of block hole m= 6 M_{sun}

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=17.7\ km

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}

R_{g}=1.1\times10^{-7}\ km

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}

R_{g}=7.4\times10^{-29}\ km

Hence, (I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

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A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
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A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

W=mg: weight of the ladder, with m = 20 kg (mass) and g=9.8 m/s^2 (acceleration of gravity)

W_M=Mg: weight of the person, with M = 54 kg (mass)

N_1: normal reaction exerted by the wall on the ladder

N_2: normal reaction exerted by the floor on the ladder

F_f = \mu N_2: force of friction between the floor and the ladder, with \mu (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}

where

Wsin 21^{\circ} is the component of the weight of the ladder perpendicular to the ladder

W_M sin 21^{\circ} is the component of the weight of the man perpendicular to the ladder

N_1 sin 69^{\circ} is the component of the normal  force perpendicular to the ladder

And solving for N_1, we find the force exerted by the wall on the ladder:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of N_2.

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

\sum F_y = 0\\N_2 - W - W_M =0

And substituting and solving for N2, we find:

N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

\sum F_x = 0\\F_f - N_1 = 0

And re-writing the equation,

\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

\mu = \frac{N_1}{N_2}

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)

And substituting, we get

N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332

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Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

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d) 1.661×10⁻²⁷KJ

Explanation:

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