Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
1 answer:
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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
Answer:
The speed the bat is gaining on its prey is 0.03m/s
Explanation:
Given;
speed of the bat, v₀ = 3.7 m/s
frequency of the bat, F₀ = 36 kHz
frequency of the source, Fs = 36.79
This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.
Apply the following equation to determine the speed of the insect which is the source;
The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s
Therefore, the speed the bat is gaining on its prey is 0.03m/s