For r > R, the electric field behaves as if all the charge were concentrated at the center, given by E = Q/(4πε₀r²).

Inside a uniformly charged spherical shell, the electric field is zero.

For a point outside the sphere, the electric field behaves as if all the charge were concentrated at the center, hence E = Q/(4πε₀r²).

As the child moves closer, the moment of inertia decreases, and to conserve angular momentum, angular velocity must increase.

Angular momentum remains constant if no external torque acts, according to the conservation of angular momentum.

As the child moves towards the center, the moment of inertia decreases, thus angular velocity increases to conserve angular momentum.

Using the formula: radius² = (distance from center to chord)² + (half of chord length)². Here, radius² = 6² + (16/2)² = 36 + 64 = 100, so radius = √100 = 10.

Using the formula d = √(r^2 - (c/2)^2), where r = 10 and c = 16, we get d = √(10^2 - 8^2) = √(100 - 64) = √36 = 6.

Circumference = 2πr; 31.4 = 2 * π * r; r = 31.4 / (2 * π) = 5 cm.

Using the formula C = πd, we find d = C/π = 62.8/3.14 = 20 cm.

Circumference = 2πr; 62.8 = 2 * π * r; r = 62.8 / (2 * π) = 10 cm.

Circumference C = πd = π(10) = 10π.

The length of an arc is given by L = (θ/360) * 2πr. Here, L = (90/360) * 2π(10) = 17.5 cm.

Diameter = 2 * radius = 2 * 10 m = 20 m.

Circumference = 2πr = 2 * 3.14 * 3 = 18.84 cm.

Circumference = 2πr = 2 * 3.14 * 4 cm = 25.12 cm.

Diameter = 2 * radius = 2 * 4 cm = 8 cm.

The area of a circle is calculated using A = πr². Thus, A = π(7)² = 49π ≈ 154 cm².

The circumference of a circle is given by 2πr. Here, r = 7 cm, so circumference = 2π * 7 = 14π cm.

Circumference = 2πr = 2 × (22/7) × 7 = 44 cm.

Circumference = 2 * π * radius = 2 * 3.14 * 7 ≈ 43.96 cm.

Area = π * r^2 = π * 7^2 = 154 (using π ≈ 22/7).

Area = πr², so r² = 50. Thus, r = √50 = 5√2 cm. Diameter = 2r = 10√2 cm ≈ 14.14 cm.

Rearranging the equation to standard form gives (x - 2)² + (y + 3)² = 0, thus the center is (2, -3).

The equation of the circle is x² + y² = 25. The point (6, 0) gives 6² + 0² = 36, which is greater than 25, hence it lies outside.

Area = πr². If radius is doubled (2r), new area = π(2r)² = 4πr², which is quadruple the original area.

Area = πr²; If r is tripled, new area = π(3r)² = 9πr²; Factor = 9.

Area = πr². If radius is tripled, new area = π(3r)² = 9πr², which is 9 times the original area.

Area = πr². If radius is tripled, new area = π(3r)² = 9πr², which is 9 times the original area.

Total resistance R = R1 + R2 = 3Ω + 9Ω = 12Ω. Using Ohm's law, I = V/R = 12V/12Ω = 1A.