Answer:
0.000311 m³
Explanation:
The formula that connect volume, mass and density is given below
D = m/v ...................... Equation 1
Where D = Density of the gold, m = mass of the gold, v = volume of the gold.
make v the subject of the equation
v = m/D .................. Equation 2
Given: m = 6 kg,
Note the density of pure gold = 19300 kg/m².
Substitute into equation 2
v = 6/19300
v = 0.000311 m³
Hence the volume that will verify that it's a pure gold = 0.000311 m³
Answer:
bar one:constant acceleration
bar two:remains stationary
bar three:constant acceleration
Answer: 1,350cm³
Explanation:
Volume = LWH
= 30 × 15 × 3 = <u>1</u>,<u>350cm</u>³
Answer:
For outer points of shell
![E = \frac{k_eQ}{r^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bk_eQ%7D%7Br%5E2%7D)
Now for inner point of shell
![E = 0](https://tex.z-dn.net/?f=E%20%3D%200)
Explanation:
As we know that out side the shell electric potential is given as
![V = \frac{K_e Q}{r}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BK_e%20Q%7D%7Br%7D)
inside the shell the electric potential is given as
![V = \frac{K_e Q}{R}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BK_e%20Q%7D%7BR%7D)
now we know the relation between electric potential and electric field as
![E = - \frac{dV}{dr}](https://tex.z-dn.net/?f=E%20%3D%20-%20%5Cfrac%7BdV%7D%7Bdr%7D)
so we can say for outer points of the shell
![E = -\frac{dV}{dr}](https://tex.z-dn.net/?f=E%20%3D%20-%5Cfrac%7BdV%7D%7Bdr%7D)
![E = - \frac{d}{dr}(\frac{K_eQ}{r})](https://tex.z-dn.net/?f=E%20%3D%20-%20%5Cfrac%7Bd%7D%7Bdr%7D%28%5Cfrac%7BK_eQ%7D%7Br%7D%29)
![E = \frac{k_eQ}{r^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bk_eQ%7D%7Br%5E2%7D)
Now for inner point again we can use the same
![E = - \frac{dV}{dr}](https://tex.z-dn.net/?f=E%20%3D%20-%20%5Cfrac%7BdV%7D%7Bdr%7D)
![E = - \frac{d}{dr}(\frac{K_eQ}{R})](https://tex.z-dn.net/?f=E%20%3D%20-%20%5Cfrac%7Bd%7D%7Bdr%7D%28%5Cfrac%7BK_eQ%7D%7BR%7D%29)
![E = 0](https://tex.z-dn.net/?f=E%20%3D%200)