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Bess [88]
3 years ago
11

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

us our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 x 10°years.
Physics
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

a) 0.0288 grams

b) 2.6*10^{-10} J/kg

Explanation:

Given that:

A typical human  body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance  for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = \frac{energy \ absorbed }{mass \ of \ the \ body}

= \frac{1.10*10^6*1.6*10^{-14}}{80}

= 2.2*10^{-10} \ J/kg

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) =  1.2  *2.2*10^{-10} \ J/kg

Effective dose (Sv) = 2.6*10^{-10} J/kg

 

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Answer:

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Explanation:

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The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele
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Answer:

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Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

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553^2=0^2+2\times a\times 0.855

a=178835.6m/sec^2

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3 years ago
A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
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Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

D=\sqrt{9^2+12^2} \\\\D=15\ km

For direction,

\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}

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2 years ago
The y-component of the force F which a person exerts on the handle of the box wrench is known to be 70 lb. Determine the x-compo
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Answer:

<em>Fx = 121.24lb</em>

<em>F = 140lb</em>

Explanation:

Since we are not given the angles subtended by the force, we can assume it to be 30 degrees.

The y component of the force expressed by the formula:

Fy = Fsintheta

Given the y-component of the force F to bee 70lb

70lb =  Fsintheta

Get magnitude of the force

F = 70/sin theta

F = 70/sin 30

F = 70/0.5

F = 140lb

Get the x-component of the force

Fx = Fcos theta

Fx = 140cos 30

Fx = 140(0.8660)

Fx = 1,212.4lb

<em>Hence the  x-component of the force sis 121.24lb</em>

<em></em>

<em>Note that the angle used was assumed. Other values can as well be used</em>

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Explanation:

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