Answer:
(a) 45 micro coulomb
(b) 6 micro Coulomb
Explanation:
C = 3 micro Farad = 3 x 10^-6 Farad
V = 15 V
(a) q = C x V
where, q be the charge.
q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb
(b)
V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad
q = C x V
where, q be the charge.
q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb
Acceleration occurs whenever the forces on an object are unbalanced.
It's the group of forces on the object that's either balanced or unbalanced.
There's no such thing as "an unbalanced force".
Beats are interference patterns between two tones of different frequencies. To prove the skeptic first, play the recorded audio as there are no beats in it. Now take two sound sources with different frequencies. When both sources are turned on, we hear notes that rise and fall at equal intervals. That's what's called beats.
A frequency beat occurs when two waves with different frequencies overlap, causing alternating cycles of constructive and destructive interference between the waves.
When we tap the table with our finger, then put our ear to the table, and tap the table surface as far as 30 cm from our ear. Then the sound of beats on the table will sound louder when we put our ears on the table. So, it can be concluded that solid objects can conduct sound better than air. This is because the molecules or particles of solid objects are denser than air.
Learn more about the beat's frequency at brainly.com/question/14157895
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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
