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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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Ryan swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill on him:
nexus9112 [7]

Part 1

If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position

F_g = ma

mg = m\frac{v^2}{R}

g = \frac{v^2}{R}

v = \sqrt{Rg}

given that

R = 1 m

g = 9.8 m/s^2

now from above equation we have

v = \sqrt{1(9.8)} = 3.13 m/s

Part b)

for minimum value of angular speed we will have

\omega = \frac{v}{R}

\omega = \frac{3.13}{1}

\omega = 3.13 rad/s

3 0
3 years ago
What do we mean by a penetrative pass in the game of football?​
Ivanshal [37]

Answer:

Penetration means forward passes can go through the opposition lines. Once these penetrative passes get through each line, it eliminates the line of players it broke through and leaves the player in possession closer to the opposition goal.

Explanation:

6 0
2 years ago
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
3 years ago
The main difference between obsessive-compulsive personality disorder (OCPD) and obsessive-compulsive disorder (OCD) is that ___
maksim [4K]

Answer:

B. people with OCD know their disorder is irrational

Explanation:

Got it right

6 0
2 years ago
Read 2 more answers
Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s
OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road.  You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

5 0
2 years ago
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