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3 years ago

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

Physics

1 answer:

timofeeve [1]3 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:

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

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A semi-infinite solid is

Anarel [89]

Answer:

D. infinitely extended in all directions

Explanation:

A semi infinite solid is infinitely extended in every direction. It has a single surface and can extend when heat is applied.

The body of a semi infinite solid is idealised, that is, when there is heat present, it expands in all directions to infinity. It can be used for a thick wall because its shape can be changed when subjected to different levels of heat near its surface.

It is also expands as heat is applied because its thickness is negligible.

This idealized body is used for earth, thick wall, steel piece of any shaped quenched rapidly etc indetermining variation of temperature near its surface & other surface being too far to have any impact on the region in short period of time since heat doesn’t have sufficient time to penetrate deep into body thus thickness can be neglected

8 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

marin [14]

Mary walks:
d 1 = 80 m, d 2 = 125 m, d 3 = 45 m
t = 10 minutes = 600 seconds;
Average speed:
v = ( d 1 + d 2 + d 3 ) / t
v = ( 80 m + 125 m + 45 m ) / 600 s
v = 250 m / 600 s
v = 0.4167 m/s ≈ 0.42 m/s
Answer:
E ) 0.42 meters/second

5 0

2 years ago

Wile E. Coyote stands on top of a 90-meter high cliff, looking down at the Roadrunner. If

Lina20 [59]

Answer:

33.516 kJ

Explanation:

Potential energy is given by:

PE = mgh

Where m is the mass, g is acceleration due to gravity, and h is the height. In this case:

PE = 38kg x 9.8m/s^2 x 90m = 33516 kg m^2/s^2 = 33516 J = 33.516 kJ

6 0

3 years ago

Define output work and input work

Olenka [21]

Answer: Input work is the work done on a machine as the input force acts through the input distance.Other ways it would be the work done on a body or system, that is, forces that are applied to the body or system. This is in contrast to output work which is a force that is applied by the body or system to something else.Output work is the work done by a machine as the output force acts through the output distance. The machine does to the object to increase the output distance.

Explanation:

6 0

3 years ago