A2,000g of c-14 is left to decay radioactively. the half-life of carbon-14 is approximately 5,700 years. what percentage of the sample will remain after 17,100 years?

Answer is: 12,6% (1/8) percentage of the sample will remain.
c₀ - initial amount of C-14.
c - amount of C-14 remaining at time.
t = 5700 y.
First calculate the radioactive decay rate constant λ:
λ = 0,693 ÷ t = 0,693 ÷ 5700 y = 0,000121 1/y = 1,21·10⁻⁴ y.
c = c₀·e∧-λ·t.
c = 2000 · e∧-(0,000121 1/y · 17100 y).
c = 252 g.
ω = 252 g ÷ 2000 g = 0,126 = 12,6%.

Work done = force ×distancebut force = mass ×gravityand gravity = 10m\s force = 2×10 = 20 work done = 20×3 = 60 joules