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2 years ago

One-eighth of an original radioactive sample will remain at the end of __________ half-lives

Physics

2 answers:

polet [3.4K]2 years ago

7 0

One eighth of an original radioactive sample will remain at the end of 3 half lives.

kow [346]2 years ago

6 0

Answer:

One-eighth of an original radioactive sample will remain at the end of <u>THREE </u>half-lives

Explanation:

As we know that half life is defined as the time after which a radioactive nuclei will remain half in the number as we compare it with the initial amount present in the sample.

So here let say initially sample contains N number of radioactive atoms

so after one half life we will have "N/2" present in that sample

after 2nd half life the number of nuclei again becomes half of the remaining

$N_1 = \frac{N}{4}$

now after 3rd half life we can say

$N_2 = \frac{N}{8}$

so here correct answer will be

One-eighth of an original radioactive sample will remain at the end of <u>THREE </u>half-lives

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ruslelena [56]

Answer:

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Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

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Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

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