The only even prime number is 2, so A ∩ B = {2}.

The union A ∪ B includes all even numbers and all multiples of 3. Thus, A ∪ B = {0, 2, 3, 4, 6, 9, ...}.

A ∩ B = {2} as 2 is the only even prime number.

det(A) = (1)(4) - (2)(3) = 4 - 6 = -2.

det(A) = (1*4) - (2*3) = 4 - 6 = -2.

det(2A) = 2^2 * det(A) = 4 * (1*5 - 2*3) = 4 * (-1) = -4.

Total marbles = 4 + 6 = 10. Probability of yellow = 6/10 = 3/5.

Total marbles = 4 + 6 = 10. Probability of green marble = 4/10 = 2/5.

Power P = V^2 / R. Here, P = 9^2 / 3 = 81 / 3 = 27 W.

In this case, the reaction force at each support is half the load applied at the center.

Using the formula for minimum deviation, D = (n - 1)A. Here, n = √3 and A = 60 degrees. Thus, D = (√3 - 1) * 60 degrees, which approximates to 30 degrees.

Total days = 3 readings * 5 days per reading = 15 days.

Total days = 5 bills * 15 days/bill = 75 days.

Applying a force at an angle reduces the normal force because part of the applied force acts vertically upward, counteracting the weight of the block.

Speed upstream = distance/time = 20/4 = 5 km/h. Speed of current = speed of boat - speed upstream = 10 - 5 = 5 km/h.

Speed upstream = speed of boat - speed of current = 15 - 5 = 10 km/h.

Angular acceleration α = τ/I = 5 N m / 15 kg m² = 0.33 rad/s².

A body moving in a circular path experiences centripetal acceleration directed towards the center of the circle.

A body moving in a circular path with constant speed experiences a centripetal force directed towards the center of the circle.

Angular velocity (ω) = 2π/T = 2π/5 = 1.25 rad/s.

The body experiences centripetal acceleration directed towards the center of the circle.

Angular momentum remains constant if no external torque acts.

Angular momentum L = Iω; if I is halved and ω remains constant, L remains L.

The body experiences centripetal acceleration directed towards the center of the circle.

Total balls = 3 + 2 = 5. Probability of selecting a red ball = Number of red balls / Total balls = 3/5.

The charge (Q) on a capacitor is given by Q = C * V. Here, Q = 10 µF * 5 V = 50 µC.

When a charged capacitor is short-circuited, the charge flows through the circuit, and the charge on the capacitor becomes zero.

Energy = 0.5 × C × V² = 0.5 × 6 × 10^-6 × 12² = 0.432 mJ.

When the voltage is increased to 10 V, the charge on the capacitor increases because charge (Q) is directly proportional to voltage (V) for a given capacitance.

When connected in parallel, charge redistributes, and the final voltage across both capacitors will be V/2.