The angular momentum of a wheel changes from \(2 \mathrm{~L}\) to $5 \mathrm{~L}$ in 3 seconds what is the magnitudes of torque acting on it? \(\\{\mathrm{A}\\} \mathrm{L}\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} \mathrm{L} / 3\) \(\\{\mathrm{D}\\} \mathrm{L} / 5\)

Short Answer

Expert verified
The correct answer is: \(\boxed{\mathrm{(A)}\, \mathrm{L}}\).

Step by step solution

01

Determine the Change in Angular Momentum

To find the change in angular momentum, we will subtract the initial angular momentum from the final angular momentum: ΔL = L_final - L_initial = 5L - 2L = 3L
02

Calculate Torque

Now that we have the change in angular momentum, we can use the formula for torque to find the magnitude of the torque acting on the wheel: Torque = ΔL / Δt = 3L / 3s = L The magnitude of the torque acting on the wheel is L. The correct answer is: \(\boxed{\mathrm{(A)}\, \mathrm{L}}\).

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If distance of the earth becomes three times that of the present distance from the sun then number of days in one year will be .... \(\\{\mathrm{A}\\}[365 \times 3]\) \(\\{\mathrm{B}\\}[365 \times 27]\) \(\\{\mathrm{C}\\}[365 \times(3 \sqrt{3})]\) \(\\{\mathrm{D}\\}[365 /(3 \sqrt{3})]\)

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