The dimension of K is M/ T^2
according to the question T=2π square root ofm/k here 2 pi is constant so
T= root of m /k and root of k = root of m/ T now by squaring on both the sides we get the answer k= M/ T^2
complete question :
A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T=√2πm/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?
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Answer:
μ =5.40 A-m²
Explanation:
The components of the net magnetic field are the magnetic field of the dipole and the magnetic field of Earth, then from the right triangle, the deflection angle is computed by
tan θ = Bdipole / Bearth ⇒ Bdipole = Bearth* tan θ
Bdipole = 2e-5 T*tan 70° = 5.49e-5 T
The magnetic field at the location of the compass due to the dipole has a magnitude
Bdipole = (μ₀/4π)(2μ/r³) ⇒ μ = Bdipole r³ / 2(μ₀/4π)
μ = (5.49e-5 T)(0.27m)³ / 2(1 × 10−7 T m² /(C m/s)) = 5.40 A-m²
T = 3.5 secs
Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
