The time period of a simple pendulum is given by T=2πlg+kxT = 2\pi\sqrt{\frac{l}{g}} + kxT=2πgl+kx, where lll is the length, ggg is acceleration due to gravity, and xxx is displacement. The dimensions of kkk are:
[L⁻¹T]
[LT⁻¹]
[L⁻¹T⁻¹]
[L⁻²T²]
Related Questions
One is equivalent to 931 MeV energy. The rest mass of electron is 9.1imes10−31kg9.1 imes {10^{ - 31}}kg9.1imes10−31kg. the Mass equivalent energy is (1 amu = =1.67imes10−17kg = 1.67 imes {10^{ - 17}}kg=1.67imes10−17kg)
0.5073 MeV
0.693 MeV
4.0093 MeV
None of these