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
If it's a distance graph, then it's a constant speed.
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + 
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
No. Oxygen( an atmosphere to contain the oxygen),water ,sunlight(energy)
A=(vf-vi)/t
a=(50-25)/10
a=2.5m/s^2
