Wavelength λ = v/f = 1500 m/s / 300 Hz = 5 m.

K.E. = 0.5 * m * v², percentage error = 2 * (Δv/v) = 2 * (0.5/20) = 0.05 or 5%.

The absolute error is given directly as ±0.5 m/s.

Percentage error = (Absolute error / Measured value) * 100 = (2 / 60) * 100 = 3.33%

Kinetic energy = 0.5 * m * v²; uncertainty in KE = 2 * v * uncertainty in v = 2 * 30 * 0.5 = 30 J.

The total mechanical energy at the top is purely potential energy, which is mgh.

Using conservation of energy, the potential energy at height h converts to kinetic energy at the bottom, giving speed √(2gh).

The total kinetic energy of a rolling sphere is the sum of translational and rotational kinetic energy, which is (1/2)mv^2 + (2/5)mv^2.

The angular speed ω of a rolling sphere is given by ω = v/R.

The electric potential inside a charged spherical conductor is constant and equal to the potential on its surface, which is Q/(4πε₀R).

The electric potential V on the surface of a charged spherical conductor is given by V = kQ/R.

The electric field outside a spherical charge distribution is given by E = Q/4πε₀r². At 2R, it becomes Q/4πε₀(2R)².

For a spherical shell, the electric field outside the shell behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).

By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.

By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.

According to Hooke's law, elongation is inversely proportional to the spring constant; thus, if the spring constant is doubled, the elongation halves.

Using Hooke's law, k = F/x = 10 N / 0.05 m = 200 N/m.

Potential Energy = 0.5 × k × x² = 0.5 × 200 N/m × (0.1 m)² = 1 J.

Work done = 0.5 × k × x^2 = 0.5 × 200 N/m × (0.1 m)^2 = 1 J.

Potential Energy in spring = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.

Work done = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.

Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.

Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.

Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(2²) = 19.6 m.

Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(3²) = 44.1 m.

Time to fall (t) = √(2h/g) = √(2*45/9.8) ≈ 3 s; horizontal distance = v*t = 10*3 = 30 m.

Time to fall t = √(2h/g) = √(90/9.8) ≈ 4.27 s; horizontal distance = 10 * 4.27 ≈ 42.7 m.

Using h = (1/2)gt², we have 45 = (1/2)(9.8)t², solving gives t ≈ 4.3 s.

Time (t) = √(2h/g) = √(2*80/9.8) = 4.04 s.

Using h = (1/2)gt², we get t = √(2h/g) = √(2*80/9.8) = 4.04 s.