A copper rod is sliding on two conducting rails that make an angle of 19o with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.63-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 7.5-s period after the rod has passed point A.

0.2923 V

Explanation:

Given that

Angle between the rails, θ = 19°

Speed of the rod, v = 0.6 m/s

Magnetic field present, B = 0.63 T

Time used, t = 7.5 s

E = -ΔΦ/Δt

where, Φ = BA, so

E = -BΔA / Δt

To get the area, if we assume the rails are joined in a triangular fashion(see attachment)

E = -B(1/2 * AC * BC) / Δt

E = -B(vΔt * vΔt tanθ) / 2Δt

E = -(B * v² * Δt² * tanθ) / 2Δt

E = -Bv²Δt.tanθ/2

E = -(0.63 * 0.6² * 7.5 * tan 19) / 2

E = -0.5857 / 2

E = -0.2923

Thus, the magnitude of average emf induced if 0.2923 V

49.5 m/s

Explanation:

Let's use the energy approach to solve this problem.

We are given the mass and displacement of the boulder.

The formula for potential energy is PE = mgh.

The formula for kinetic energy is KE = 1/2mv².

If we ignore air resistance, we can use the conservation of energy to solve this problem by setting PE and KE equal to each other. By doing so, we realize that the mass of the boulder does not affect the speed that it hits the ground (final velocity).

Divide both sides of the equation by m.

Let's take the downwards direction as positive. The displacement of the boulder is 125 m and g = 9.8 m/s².

Now we can solve for v.

Multiply both sides by 2.

Take the square root of both sides.

Rounded to the nearest 0.1 m/s, the final velocity of the boulder is 49.5 m/s.