Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is
Explanation:
From the question we are told that
The radius of the inner circle is 
The radius of the outer circle is 
The charge on the spherical shell
The magnitude of the point charge at the center is 
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

substituting values

where k is the coulomb constant with value 
substituting values


Answer: f = 927.55Hz
Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below
L = λ/4 where λ = wavelength.
speed of sound in air = v = 343m/s.
fundamental frequency of open closed tube = 315Hz
λ = 4L.
v = fλ
343 = 315 * 4L
343 = 1260 * L
L = 343/ 1260
L = 0.27m
In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.
The length of tube and wavelength are related by the formulae below
L = λ/4, λ=4L
λ = 4 * 0.27
λ = 1.087m.
v = fλ
1010 = f * 1.087
f = 1010/1.807
f = 927.55Hz
Answer:
(a) 
(b) 
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

(a) The electromagnetic force of the battery is
and its internal resistance is
. Knowing the equivalent resistance of the headlights is
, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

Solving for i

i=2.28\ A
The potential difference across the headlight bulbs is


(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is
