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umka2103 [35]
3 years ago
6

What does Newton's third law describe?

Physics
1 answer:
Dennis_Churaev [7]3 years ago
7 0
A. Equal and opposite forces

Explanation:
History law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exert an equal and opposite force on object A
You might be interested in
What elements had to be made in a supernova explosion?
hram777 [196]

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.

4 0
3 years ago
Which is the BEST example of the kind of mechanics that are studied in sports biometrics?
ikadub [295]

Answer:

-A.

Explanation:

: Hope it's Help:

[correct me if I'm not correct]

5 0
1 year ago
1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i
ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A  is m_a=2\ kg

mass of block B is m_b=3\ kg

The initial velocity of block A u_a=5\ m/s

the initial velocity of block B is u_b=0

After collision velocity of block A is v_a=1\ m/s

Conserving momentum

m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s

The momentum of block A after the collision is P_a=2\times 1=2\ kg.m/s

Therefore, there is no change in sign.

6 0
2 years ago
A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul
Marina CMI [18]
The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. 
5 0
2 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
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