PLZ HELP do waves transport water with them
1 answer:
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Answer:
(A) F
Explanation:
See it in the pic.
Answer:
(a) The potential difference between the spheres is 750 kVA
(b) The charge on the smaller sphere is 6. μC
(c) The charge on the smaller sphere, Q₁ = 13. μC
Explanation:
(a) The given parameters are;
The charge on the inner sphere, q = 5.00 μC
The radius of the inner sphere, r = 3.00 cm = 0.03 m
The charge on the larger sphere, Q = 15.0 μμC
The radius of the larger sphere, R = 6.00 cm = 0.06 m
The potential difference between two concentric spheres is given according to the following equation;
Where;
R = The radius of the larger sphere = 0.06 m
r = The radius of the inner sphere = 0.03 m
q = The charge of the inner sphere = 5.00 × 10⁻⁶ C
Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C
k = 9 × 10⁹ N·m²/C²
Therefore, by plugging in the value of the variables, we have;
The potential difference between the spheres, = 750,000 N·m/C = 750 kVA
(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;
= Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C
= 20 × 10⁻⁶ C
From which we have;
Q₁/Q₂ = R/r
Where;
Q₁ = The new charge on the on the larger sphere
Q₂ = The new charge on the on the smaller sphere
= 20 × 10⁻⁶ C = Q₁ + Q₂
∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂
∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2
20 μC - Q₂ = 2·Q₂
20 μC = 3·Q₂
Q₂ = 20 μC/3
The charge on the smaller sphere, Q₂ = 20 μC/3 = 6. μC
(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3
The charge on the smaller sphere, Q₁ = 40 μC/3 = 13. μC.
Answer and Explanation:
We know that resistance from the given equation of resistance it is clear that resistance depends on resistivity length and area of the material but we can not change the length because it is given that the length must be 2.5 cm long.
So we can do two two things to reduce the resistance
- increase the cross sectional area
- decrease the resistivity of the material