Answer:
Half life is 6 years.
Explanation:
T½ = In2 / λ
Where λ = decay constant.
But N = No * e^-λt
Where N = final mass after a certain period of time
No = initial mass
T = time
N = 0.625g
No = 10g
t = 24 years
N = No* e^-λt
N / No = e^-λt
λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)
λ = -(1 / 24) * In (0.625/10)
λ = -0.04167 * In(0.0625)
λ = -0.04167 * (-2.77)
λ = 0.1154
T½ = In2 / λ
T½ = 0.693 / 0.1154
T½ = 6.00 years.
The half life of radioactive cobalt-60 is 6 years
Explanation:
The volumetric flow rate of water will be as follows.
q = 
= 0.0378 
Diameter =
= 0.2032 m
Relation between area and diameter is as follows.
A =
=
= 0.785 x 0.2032 x 0.2032
= 0.0324
Also, q = A × V
or, V = 
= 
= 1.166 m/s
As, viscosity of water = 1 cP =
Pa-s
Density of water = 1000
Therefore, we will calculate Reynolds number as follows.
Reynolds number =
=
= 236931.2
Hence, the flow will be turbulent in nature.
Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.