<span>Cobalt-60 is undergoing a radioactivity decay.
The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span> n </span>⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.
So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567
t = 3÷0.567
= 5.290626524
the half-life of Cobalt-60 is 5.29 years.
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It might be pull at a force of 100 N. I might be wrong.
Answer:
7174.09 N
Explanation:
Area = 7 ft x 4 x ft = 28 ft^2 = 2.60129 m^2
Pressure = 0.4 psi = 2757.9 Pa
Force = Pressure x Area
Force = 2757.9 x 2.60129 = 7174.09 N
Mass is given by:
m = ρV
m = mass, ρ = density, V = volume
Given values:
ρ = 1.11kg/m³, V = 350m³
Plug in and solve for m:
m = 1.11(350)
m = 389kg