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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

Jupiter i hope it is right answer
Answer:
<u>Facts about 258</u>
<u>Sig Figs</u>
3
<u>258</u>
<u>Decimals</u>
0
<u>Scientific Notation</u>
2.58 × 102
<u>E-Notation</u>
2.58e+2
<u>Words</u>
two hundred fifty-eight
Explanation:
258 Rounded to Fewer Sig Figs
2 260 2.6 × 102
1 300 3 × 102
<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here
then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))
The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>