Hello!
A half-life is the amount of time it takes half of an element to decay. For example, after one of carbon's half-life, half of carbon would've decayed.
Now, we can calculate how much has decayed if 5 half lives have passed.
1 -> 50% (1) -> 25% (2) -> 12.5% (3) -> 6.25% (4) -> 3.125% (5)
Therefore, there would be 3.125% left of iodine. However, your question is asking what percent has decayed. Therefore, subtract 3.125 from 100% to get your final answer.
100 - 3.125 = 96.875
Therefore, your final answer is 96.875% of iodine has decayed if 5 half lives have passed.
Hope this helps!
I think the correct answer from the choices listed above is option C. The energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface by the fact that o<span>rganisms in the deep sea do not have direct access to sunlight. Hope this answers the question. Have a nice day.</span>
Ten is your answer I think
The correct answers of this question is options A and C. Types of context clues are direct explanation or variation, and restatement or synonym. Context clues are hints that a writer gives to explain a difficult word. Other types are antonyms, interference and punctuation.
Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
