Increasing capacitance increases the current in a capacitive AC circuit, as current I = V/Xc.

Increasing the current in a coil increases the magnetic field produced by the coil.

Increasing the current in a solenoid increases its magnetic field strength, as the magnetic field is directly proportional to the current.

Capacitive reactance Xc = 1/(2πfC), so increasing frequency decreases Xc.

The capacitive reactance decreases with increasing frequency, thus decreasing the impedance.

The impedance of a capacitor decreases with increasing frequency.

Inductive reactance Xl = ωL increases with frequency since Xl = 2πfL.

The total impedance can increase or decrease depending on the relationship between R, XL, and XC as frequency changes.

Increasing the number of turns in a coil increases the magnetic field strength produced by it, as the field strength is directly proportional to the number of turns.

Increasing the slit width decreases the fringe width because the angle of diffraction decreases.

Increasing temperature generally decreases the magnetism of ferromagnetic materials, as thermal agitation disrupts the alignment of magnetic domains.

Fringe separation is directly proportional to the distance from the slits to the screen. Increasing this distance increases the fringe separation.

Removing a resistor from a series circuit decreases the total resistance since the sum of the remaining resistances is less.

Efficiency = 1 - (T_c/T_h) = 1 - (300/600) = 0.5.

Efficiency = 1 - (T2/T1) = 1 - (300/500) = 0.4 or 40%.

Efficiency (η) = 1 - (T_c/T_h) = 1 - (300/600) = 0.5 or 50%.

Efficiency (η) = 1 - (T2/T1) = 1 - (300/500) = 0.4 or 40%.

The electric field (E) due to a point charge (q) at a distance (r) is given by E = k * q / r^2, where k is Coulomb's constant.

E = k * |q| / r^2 = (9 × 10^9) * (3 × 10^-6) / (0.3)^2 = 90 N/C.

V = k * q / r = (9 × 10^9) * (5 × 10^-6) / 0.4 = 112.5 V.

The electric potential (V) at a distance r from a point charge Q is given by V = kQ/r.

Energy = 13.6 eV * (1/n1^2 - 1/n2^2) = 13.6 * (1/2^2 - 1/3^2) = 3.40 eV.

Using the formula 1/R_eq = 1/R1 + 1/R2, we get 1/R_eq = 1/4 + 1/4 = 1/2, thus R_eq = 2Ω.

Using the formula 1/R_eq = 1/R1 + 1/R2, we get 1/R_eq = 1/4 + 1/12 = 1/3, so R_eq = 3Ω.

Escape velocity = √(2GM/R) = √(2 * 6.67 × 10^-11 * 6 × 10^24 / 4 × 10^6) = 11,200 m/s

Escape velocity (v) = √(2gR) = √(2 * 9.8 * 6.4 × 10^6) ≈ 11.2 km/s

Escape velocity v = √(2gR) = √(2 * 9.8 * 6.4 × 10^6) ≈ 11.2 km/s.

Using heat balance: m_ice * L_f + m_water * c * (T_final - 80) = 0. Solving gives T_final = 20°C.

Using the principle of conservation of energy, the heat lost by water equals the heat gained by ice. Solving gives a final temperature of 20°C.

Using Coulomb's law, F = k * |q1 * q2| / r^2 = (9 × 10^9) * (2 × 10^-6) * (3 × 10^-6) / (0.5)^2 = 0.24 N.