Why is the astronomical unit used to measure distances in the solar system?
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Answer:
a) At 0.20 m, the magnitude of the field is 675.0 kV
The direction of the field is acting outwards from the center of the charged spheres
b) At 0.10 m, the magnitude of the field is 135 kV
The direction is acting outwards from the center of the charged spheres
c) At 0.025 m
The magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres
Explanation:
The charged spherical shell parameters are;
The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C
The radius of the inner shell, R₁ = 0.050 m
The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C
The radius of the outer shell, R₂ = 0.15 m
Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;
When r < R₁ < R₂,
When R₁ < r < R₂,
When R₁ < R₂ < r,
a) When r = 0.20 m, we have;
R₁ < R₂ < r, therefore
By plugging in the values, we get;
Therefore, the magnitude of the field, V = 675.0 kV
The direction of the field is outwards
b) When r = 0.10 m, we have;
When R₁ < r < R₂, therefore;
By plugging in the values, we get;
Therefore, the magnitude of the field, V = 135 kV
The direction of the field is outwards from the center
c) When r = 0.025 m, we have;
When r < R₁ < R₂, therefore;
By plugging in the values, we get;
Therefore, the magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres.
Answer:
decreases
Explanation:
The angular frequency of the mass is given by
where, k is the spring constant and m be the mass of the body which is doing oscillations.
if the mass of the body increases, so the value of angular frequency decreases, as the angular frequency is inversely proportional to the square root of the mass of the body.