If Hubble's constant is 73 km/s/Mpc then the age of the universe is 13.4 billion years, assuming a uniform expansion. Suppose it
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Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
The approximate radius of the nucleus of this atom is 4.656 fermi.
b)
k= = Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
Force of repulsion between two protons on opposite sides of the diameter
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is .
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as
Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is
Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,
Part(c):
If we apply Gauss' law of electrostatics, then