Atoms found in nature are either stable or unstable. An atom is stable if the forces among the particles that makeup the nucleus are balanced. An atom is unstable (radioactive) if these forces are unbalanced; if the nucleus has an excess of internal energy. Instability of an atom's nucleus may result from an excess of either neutrons or protons. Ccredits: internet source
To solve this problem we will begin by applying the given relations of density in terms of mass and volume, and from this last value we will take its geometric measurement for a sphere (Approximation of a planet) From there we will find the radius of the planet. Finally we will make a comparison between the radius of the new planet and the radius of the earth to understand its proportion.
Defining the Volume variables we have to
![V = \frac{m}{\rho}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bm%7D%7B%5Crho%7D)
Here
V= Volume
m = mass
=Density
For a spherical object the Volume is
![V = \frac{4}{3} \pi R^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20R%5E3)
PART A)
Equation we have
![\frac{4}{3} \pi r^3 = \frac{m}{\rho}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E3%20%3D%20%5Cfrac%7Bm%7D%7B%5Crho%7D)
![R = (\frac{3}{4\pi}(\frac{m}{\rho}))^{1/3}](https://tex.z-dn.net/?f=R%20%3D%20%28%5Cfrac%7B3%7D%7B4%5Cpi%7D%28%5Cfrac%7Bm%7D%7B%5Crho%7D%29%29%5E%7B1%2F3%7D)
In this case the mass of new planet is 5.5times the mass of Earth,
![m = 5.5m_E](https://tex.z-dn.net/?f=m%20%3D%205.5m_E)
Then,
![R = (\frac{3}{4\pi}(\frac{5.5m_E}{\rho}))^{1/3}](https://tex.z-dn.net/?f=R%20%3D%20%28%5Cfrac%7B3%7D%7B4%5Cpi%7D%28%5Cfrac%7B5.5m_E%7D%7B%5Crho%7D%29%29%5E%7B1%2F3%7D)
The mass of the Earth is
kg and the density is
,
Replacing we have that,
![R = (\frac{3}{4\pi}(\frac{5.5(5.97*10^{24})}{1.76}))^{1/3}](https://tex.z-dn.net/?f=R%20%3D%20%28%5Cfrac%7B3%7D%7B4%5Cpi%7D%28%5Cfrac%7B5.5%285.97%2A10%5E%7B24%7D%29%7D%7B1.76%7D%29%29%5E%7B1%2F3%7D)
![R = 1.64*10^7m](https://tex.z-dn.net/?f=R%20%3D%201.64%2A10%5E7m)
![R = 1.64*10^4 Km](https://tex.z-dn.net/?f=R%20%3D%201.64%2A10%5E4%20Km)
Therefore the radius of this new planet is ![R = 1.64*10^4 Km](https://tex.z-dn.net/?f=R%20%3D%201.64%2A10%5E4%20Km)
PART B) The value of radius of the Earth is
![R_E = 6.37*10^3 km](https://tex.z-dn.net/?f=R_E%20%3D%206.37%2A10%5E3%20km)
Then the relation between them is
![\frac{R}{R_E}= \frac{1.64*10^4}{6.37*10^3}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR_E%7D%3D%20%5Cfrac%7B1.64%2A10%5E4%7D%7B6.37%2A10%5E3%7D)
![\frac{R}{R_E}= 2.57](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR_E%7D%3D%202.57)
Therefore the radius of the new planet in terms of radius of the Earth is ![2.57R_E](https://tex.z-dn.net/?f=2.57R_E)
Answer:
The answer is c
Thermal energy moves within the air from the flames to the marshmallow.
Explanation:
Hope it helps
'D. 18 feet' sounds right
Most Likely Answer:
It cannot be determined without knowing the mass of the sphere.