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
V o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².
Explanation:
It is given that, the height of a certain tower is 862 feet i.e to reach on the ground the object should travel, s = 862 feet
The distance traveled by a freely falling object is given by :
t = 7.34 seconds
So, the object will take 7.34 seconds to fall to the ground from the top of the building. Hence, this is the required solution.
Spherical because it’s more like clouds
Answer:
r = 1.45 Å
Explanation:
given,
λ = 1.436 Å
θ = 20.62°
d = a
n = 2
metal gold crystallizes in a face centered cubic unit cell
Radius of the gold atom = ?
using Bragg's Law
n λ = 2 d sin θ
2 x 1.436 Å = 2 a sin 20.62°
a = 4.077 Å
We know relation of radius for face centered cubic unit cell
r = 1.45 Å
the radius of a(n) gold atom. is equal to 1.45 Å
