The kinetic energy before equals K after
Answer:
Ro = 8.65 [g/cm³]
Explanation:
We must remember that density is defined as the ratio of mass to volume.

where:
m = mass = 0.450 [kg] = 450 [g]
V = volumen = 52 [cm³]
Ro = density [g/cm³]
Now replacing:
![Ro = 450/52\\Ro = 8.65 [g/cm^{3} ]](https://tex.z-dn.net/?f=Ro%20%3D%20450%2F52%5C%5CRo%20%3D%208.65%20%5Bg%2Fcm%5E%7B3%7D%20%5D)
Answer:
P=(2 nm, 8mn)
Explanation:
Given :
Position of positively charged particle at origin, 
Position of desired magnetic field, 
Magnitude of desired magnetic field, 
Let q be the positive charge magnitude placed at origin.
<u>We know the distance between the two Cartesian points is given as:</u>

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>




as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.
assuming that the second charge is placed at (x,y) nano-meters.
Therefore,

and

Air pressure is the wi get of air molecules pressing down on the earth. The pressure of the air molecules changes as you move upward from sea level into the atmosphere, the highest pressure is at sea level where the density of the air molecules is the greatest.
Do you not understand how to solve for the answer?