A sample of cobalt-60 has an activity of 2000 Bq. What will the activity be after 10 years? Cobalt-60 has a half life of 5 years.

500.3 Bq

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/12 = half life = 5 years

Ao = initial activity = 2000 Bq

0.693/5 = 2.303/10 log (2000/A)

0.1386= 0.2303  log (2000/A)

0.1386/0.2303 =  log (2000/A)

0.6018 = log (2000/A)

(2000/A) = Antilog (0.6018)

(2000/A) = 3.9976

A = 2000/3.9976

A = 500.3 Bq

this could be newton#'s law of cooling. the cooling curve is exponential

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.