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Svetradugi [14.3K]
3 years ago
10

The life cycle of stars may be boiled down into a tug of war, lasting billions of years, between two basic forces. In your own w

ords, describe these forces, and identify which of the two eventually "wins" the battle.
Physics
1 answer:
Mumz [18]3 years ago
5 0

Answer:

Stars are very massive stellar objects, which means that they have a very intense force of gravity. This is the first of the forces entering this "war".

In addition to that, due to the force of gravity that drives the star to contract, the process known as fusion occurs (the union of atoms of one element that results in another element, hydrogen fuses in stars to produce helium). The fusion created in the high temperatures of the center of the star generates an enormous amount of energy (which causes the stars to shine) and a force going outward of the star counteracting gravity, this is the second force in the "war" .

In a stable star these two forces (gravity going inward and the pressure created by the fusion going outward ) are in balance, preventing the star from exploding or collapsing. But eventually the star exhausts its "fuel" (hydrogen atoms) to produce fusion within it (although stars also fuse helium and other heavier elements, but once the hydrogen is finished the star is near its end), which decreases the force outward from the star, making the force that wins this battle to be the force of gravity.

When the force of gravity wins, the star collapses on itself and from here, depending on the star's mass, several things can happen, such as the star becoming a white dwarf, a supernova, even a black hole.

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How does temperature,barometric pressure,humidity,wind speed and direction, and precipitation determine the weather in a particu
Deffense [45]
Temperature is how hot or cold something is,barometric is air pressure(when the pressures high,the weather is dry)humidity is how moist the air is and how many water particles are there, wind speed and direction is how the hot and cold air is moving and how fast,which makes wind (high to low, hot to cold) precipitation is how much it is raining at that point.
7 0
3 years ago
The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a
Mnenie [13.5K]

Answer:

19.2*10^6 s

Explanation:

The equation for time dilation is:

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.

7 0
3 years ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
3 years ago
Mercury is characterized by
cupoosta [38]
 <span>It's close to the sun without much atmosphere, so it's characterized by </span><span>very extreme temperatures.

Happy studying ^_^</span>
8 0
3 years ago
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