Answer:
There are actually three, Kepler's laws that is, of planetary motion: 1) every planet's orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet's orbital period is proportional to the cube of the semi-major axis of its
Answer:
10⁴¹ s quark top lives have been in the history of the universe.
Explanation:
You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:
![t\frac{Age of the universe}{Lifetime of a top quark}](https://tex.z-dn.net/?f=t%5Cfrac%7BAge%20of%20the%20universe%7D%7BLifetime%20of%20a%20top%20quark%7D)
Yo know that the "Age of the universe" is 100,000,000,000,000,000 which can also be expressed as 10¹⁷ s
.
You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.
Then ![t=\frac{10^{17} }{10^{-24} }](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B10%5E%7B17%7D%20%7D%7B10%5E%7B-24%7D%20%7D)
Recalling that the result of dividing two powers of the same base is another power with the same base where the exponent is the subtraction of the initial exponents, it is possible to calculate this division as follows:
![t=10^{17-(-24)}](https://tex.z-dn.net/?f=t%3D10%5E%7B17-%28-24%29%7D)
![t=10^{17+24}](https://tex.z-dn.net/?f=t%3D10%5E%7B17%2B24%7D)
<u><em>t=10⁴¹ s</em></u>
So <u><em>10⁴¹ s quark top lives have been in the history of the universe.</em></u>
Answer is 3. Volume= mass divided by density
The answer would be 5.6x10^5
Is there any answers? Or is it asking you to choose?