Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
Answer:
0.064 mg/kg/day
6.25% from water, 93.75% from fish
Explanation:
Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.
The BCF = 10³, so the concentration of the chemical in the fish is:
10³ = x / (0.1 mg/kg)
x = 100 mg/kg
For 2 L of water and 30 g of fish:
2 kg × 0.1 mg/kg = 0.2 mg
0.030 kg × 100 mg/kg = 3 mg
The total daily intake is 3.2 mg. Divided by the woman's mass of 50 kg, the dosage is:
(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day
b) The percent from the water is:
0.2 mg / 3.2 mg = 6.25%
And the percent from the fish is:
3 mg / 3.2 mg = 93.75%
Answer:
The pressure difference across hatch of the submarine is 3217.68 kpa.
Explanation:
Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.
Given:
Height of the hatch is 320 m
Surface gravity of the sea water is 1.025.
Density of water 1000 kg/m³.
Calculation:
Step1
Density of sea water is calculated as follows:

Here, density of sea water is
, surface gravity is S.G and density of water is
.
Substitute all the values in the above equation as follows:


kg/m³.
Step2
Difference in pressure is calculated as follows:


pa.
Or

kpa.
Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.