The time period (T) is the reciprocal of frequency (f). T = 1/f = 1/60 Hz = 0.0167 s.

Capacitive reactance X_C = 1/(ωC) decreases as frequency increases.

The kinetic energy of the emitted electrons is given by KE = hf - φ. If the frequency is doubled, the kinetic energy increases as it is directly proportional to frequency.

Evaluating f(0) = e^0 + 0^2 = 1 + 0 = 1.

The vertex form gives the minimum value at x = 2, f(2) = 1.

Setting f'(x) = 0 gives x = 1 as a local maximum.

A high resistance in the galvanometer increases the sensitivity of the bridge.

In a balanced Wheatstone bridge, the current through the galvanometer is zero.

A deflection in the galvanometer indicates that the bridge is unbalanced, meaning the ratios of resistances are not equal.

A deflection in the galvanometer indicates that there is a potential difference across it, meaning the bridge is unbalanced.

A deflection in the galvanometer indicates that there is a potential difference across it, meaning the bridge is not balanced.

The gravitational potential is negative and is given by V = -g * r. At infinity, V approaches 0, so at a field strength of 10 N/kg, V = -10 J/kg.

The gravitational potential V is related to the gravitational field strength g by V = -g * r. If we consider r to be 1 meter, V = -10 * 1 = -10 J/kg.

Gravitational potential V = -g * r = -10 N/kg * 2 m = -20 J/kg.

Gravitational potential V = -g * r = -10 N/kg * 2 m = -20 J/kg.

Gravitational potential V = -g * r = -10 N/kg * 2 m = -20 J/kg.

The gravitational potential V is related to the gravitational field strength g by V = -g * r. If we consider r to be 1 meter, V = -10 * 1 = -10 J/kg.

Gravitational potential V = -g * r = -10 * 2 = -20 J/kg.

Gravitational potential V = -g * h = -10 N/kg * 2 m = -20 J/kg.

Gravitational potential V = -g * r = -10 * 2 = -20 J/kg.

Using V = -g * r, we have -15 = -3 * r, thus r = 15/3 = 5 m.

Using V = -g * r, we have -20 = -4 * r, thus r = 20/4 = 5 m.

Using V = -g * r, we have -30 = -5 * r, thus r = 30/5 = 6 m.

The gravitational field strength can be found using the relation E = -dV/dr. If V = -30 J/kg, E = 3 N/kg.

The area of a triangle is given by (1/2) * base * height = (1/2) * 10 * 12 = 60 cm².

Magnification (m) = v/u = 1 when image distance equals object distance.

Using the lens formula 1/f = 1/v - 1/u, we have 1/f = 1/40 - 1/(-30) = 1/40 + 1/30. Solving gives f = 20 cm.

Inductive reactance X_L = 2πfL = 2π(50)(2) = 628 Ω.

At destructive interference, the intensity is zero because the waves cancel each other out.

Doubling the intensity of light increases the number of photons incident on the surface, which in turn increases the number of emitted electrons, assuming the frequency is above the threshold frequency.