All categories

2 years ago

Find the radius to which the sun must be compressed for it to become a black hole.

Physics

1 answer:

timofeeve [1]2 years ago

7 0

Answer:

2950 m

Explanation:

The radius to which the sun must be compressed to become a black hole is equal to the Schwarzschild radius, defined as:

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

where

G is the gravitational constant

M is the of the sun

c is the speed of light

The mass of the sun is

$M=1.99\cdot 10^{30}kg$

So, if we substitute the values of the other constants inside the formula, we find the value of the Schwarzschild radius for the sun:

$R=\frac{2(6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2})(1.99\cdot 10^{30} kg)}{(3\cdot 10^8 m/s)^2}=2950 m$

You might be interested in

A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)

Illusion [34]

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

$W=mg\Delta h$

where

m is the mass of the object

g is the acceleration of gravity

$\Delta h$ is the change in height of the object

For the object in this problem, we have:

m = 5 kg

$g=10 m/s^2$

$\Delta h = 2 m$

Substituting into the equation,

$W=(5)(10)(2)=100 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

in the figure shown, if the mass of the block is 4kg and the coefficient of static friction is 0.5 and the coefficient of kineti

Elan Coil [88]

Answer:

Ff = 19.6 N

Explanation:

So since its saying whats the minimum F to move the block, we will use static friction (0.5).

We will use the equation for force of friction, which is Ff = uFn

Ff = (0.5)(4)(9.8)

Ff = 19.6 N

this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block

3 0

3 years ago

The table below describes 100 offspring of the same two parents. What are the most likely genotypes of the parents?

mars1129 [50]

Answer:

im very con fused on what you mean by this

Explanation:

4 0

3 years ago

Read 2 more answers

A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef

inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0

2 years ago