Explanation:
The given data is as follows.
q = 6.0 nC = 
inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)
So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.
Formula to calculate the charge density is as follows.
.......... (1)
Since, area of the sphere is as follows.
A = 
........... (2)
Hence, substituting equation (2) in equation (1) as follows.
= 
= 
or, = 4.77 
Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 .
<h2>Answer: protons and neutrons.
</h2>
The atomic nuclei of almost all elements consist of protons and neutrons.
The nucleus of an atom has very small dimensions. However, it <u>occupies its central part and concentrates more than 99% of its total mass.
</u>
It is in the nucleus that the protons (positive charge) and neutrons (neutral charge) are found.
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut + 
at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) + x a x 5.2
53 = 2.6a
a = 
= 20.4m/s²
F = G*((m sub 1*m sub 2)/r^2)
