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mars1129 [50]
3 years ago
7

What elements had to be made in a supernova explosion?

Physics
1 answer:
hram777 [196]3 years ago
4 0

Answer:

Gold, Platinum and Uranium

Explanation:

A star is nothing but a huge ball of gas. Specifically, Hydrogen, the simplest element of nature. A star is in equilibrium because its immense mass causes it to collapse towards itself, squeezing those hydrogen nuclei or protons, and the union of the protons in its nucleus causes the star to explode, releasing energy. As long as these nuclear reactions exist (the same ones that human beings can cause with their hydrogen bombs), the star will remain in equilibrium.

Protons have a positive charge and tend to repel. But inside the stars they are so tight (there is a lot of pressure and temperature), that they can't avoid crashing. At that time, the electromagnetic force is defeated by what physicists call Strong Force, which holds together protons and neutrons forming more complex atoms. In a typical star, the protons join to form the next element in the periodic table: Helium, consisting of 2 protons and two neutrons. It is a rare element on Earth and was discovered in the Sun rather than on our planet. Hence his name, from the Greek Helios, the sun god.

However, the mass of the sum of the protons that bind to form Helium is less than the total mass of Helium. What happen? Are the laws of physics inside the stars not fulfilled? What happens is impossible to understand if one is born before Albert Einstein, but today it is very easy to explain. The mass that we lack, has actually become energy. The German physicist Albert Einstein (1879-1955) discovered that mass and energy are equivalent while formulating his Theory of Relativity. In fact, let me, for once, write a mathematical equation of an unparalleled beauty:

E = mc2

This equation tells us that the energy E is equal to the mass m times the square of a constant c; that constant c is the speed of light, approximately 300,000 km / s. That is, a very small mass, such as a proton, is equivalent to a very large energy, since the numerical factor by which the mass is multiplied is a very large number. And that energy is what the stars release, the one that our Sun emits and gives us life.

When Hydrogen is depleted, the star collapses until the pressure and temperature increase enough for Helium to fuse with itself; the cycle is repeated and the star ends up generating Carbon, Oxygen, Nitrogen, Silicon, Iron. As you can see, the stars are factories of atoms. When the star explodes, even heavier atoms are generated, such as Gold, Platinum, Uranium, elements that abound on our planet. And they abound, because the Sun is a second or third generation star: that is, it was born from the remains of other stars' explosions, along with the materials that make up our planet, the rest of the planets, the asteroids, the comets, the interstellar dust and ourselves.

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What pressure, in millimeters of mercury (mm Hg), is equivalent to 2.13 atmospheres
Sati [7]
The "Standard Atmosphere" is 760 mm Hg . 2.13 times that pressure is (2.13 x 760) = 1,618.8 mm Hg.
6 0
3 years ago
Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected
Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance C= 20\times10^{-2}\ F

Inductance L=20\times10^{-3}\ H

Resistance R=  50\ Omega

We need to calculate the impedance

Using formula of impedance

z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}

z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}

z=50.00

We need to calculate the phase angle

Using formula of phase angle

\theta=\cos^{-1}(\dfrac{R}{z})

\theta=\cos^{-1}(\dfrac{50}{50.00})

\theta=0.0180\ rad

Hence, The phase angle is 0.0180 rad.

3 0
3 years ago
The centripetal force acting on the space shuttle
tamaranim1 [39]

Answer:

(4) weight

Explanation:

The centripetal force acting on the space shuttle in orbit is given by:

F=m\frac{v^2}{r}

where

m is the mass of the shuttle

v is the tangential speed of the shuttle

r is the radius of its circular orbit

When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the Earth's mass

And this force, therefore, corresponds to the centripetal force.

7 0
3 years ago
Read 2 more answers
Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0
3 years ago
Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?
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<span>The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.</span>
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3 years ago
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