How Many Significant Figures for Time and Mass Calculations?

AI Thread Summary
When dividing a time of 21.76 seconds into increments for a centripetal force experiment, the result should maintain four significant figures, as the original measurement has four. For mass calculations, multiplying a mass of 2567 kg by the acceleration due to gravity (9.81 m/s²) results in an answer with three significant figures due to the precision of the values involved. If a less precise value for gravity, such as 9.7, is used, the final answer would only have two significant figures. The discussion emphasizes the importance of considering the precision of all measurements when determining significant figures in calculations. Accurate reporting of significant figures is crucial for maintaining the integrity of scientific data.
helpmedude
Hi just a quick question about sig figs. If one was too divide a time such as 21.76s into increments of twenty to find the period of revolution for a centripetal force experiment would the answer be two or four significant figures? also if I multiply a mass such as 2567kg by acceleration due to gravity should my final answer be two significant figures? I am sure the answer to my second question is yes but the first I am not so sure about.
 
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21.76 is 4 sig figures. If you measured say 20 rotations to get this answer then yes, you know the time of one rotation to 4sig figures.

If you have a mass of 2567kg (again to 4 sig fig) and divide by g=9.81 then you only have 3sig figures in you answer. Or more correctly you have mass of 2567 +- 0.5kg and g=9.81 +- 0.005 so you know g to 1 part in 2000 you would know your answer to 1 part in 2000, ie approx +- 1kg.
If you used g=9.7 then you have only 2sig figures and an error of 1 part in 100.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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