(2) According to the simple pendulum period formula T=2πlg, the acceleration of gravity is g=4π2lT2.
A, the measured value of gravity acceleration has nothing to do with the amplitude, and the amplitude is too small, which does not affect the measurement result, so A is wrong;
B when the initial timing error is recorded as n= 1, the period is too small. According to g=4π2lT2, G is too large, so B is correct;
C, take the sum of the cycloid length and the diameter of the pendulum ball as the pendulum length, and the pendulum length is too large. According to g=4π2lT2, G is too large, so C is correct;
D, before hanging pendulum ball, measure the pendulum length, pendulum length is too small. It can be seen from g=4π2lT2 that G is too small, so D is wrong;
So choose BC
(3) The difference of gravity acceleration measured by this classmate △g=g0-g is caused by the density of spherical minerals being greater than that of the earth.
It is equivalent to a spherical object with a density of (nρ-ρ) and a volume of V, which produces a gravitational acceleration △ g on the earth's surface with a distance of R. 。
So there is: △g=G(nρ? ρ)Vr2
r2=G(nρ？ ρ)V△g=Gρ(n？ 1)Vgg0
The solution is: r=Gρ(n? 1)Vgg0
Therefore, the answer is: (1) 2.25 s; (2) BC; (3) I praised it and stepped on it. What is your evaluation of this answer? Review and prospect of SLD laboratory technology
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