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3 years ago

15

A student starts a food fight by throwing a 0.5 kg burrito at some girl he likes. He throws it kind-of hard so he accelerates it

at 3 m/s2. How much force did this take?

1 answer:

Elenna [48]3 years ago

4 0

Explanation:

m = mass of burrito thrown by the student = 0.5 kg

a = acceleration of the burrito thrown by the student = 3 m/s²

F = force applied by the student on the burrito = ?

According to newton's second law , the net force on an object is the product of its mass and acceleration. it is given as

F = ma

inserting the values

F = (0.5) (3)

F = 1.5 N

hence the net force on the burrito comes out to be 1.5 N

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Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus 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 $1.28 * 10^9$ years.

Answer:

The potassium-40 present in 80 kg is $Z = 0.0288 *10^{-3}\ kg$

The effective dose absorbed per year is $x = 2.06 *10^{-24}$ per year

Explanation:

From the question we are told that

The mass of potassium in 1 kg of human body is $m = 3g= \frac{3}{1000} = 3*10^{-3} \ kg$

The mass of the person is $M = 80 \ kg$

The abundance of Potassium-39 is 93.26%

The abundance of Potassium-40 is 0.012%

The abundance of Potassium-41 is 6.78 %

The energy absorbed is $E = 1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J$

Now 1 kg of human body contains $3.0*10^{-3}\ kg$ of Potassium

So 80 kg of human body contains k kg of Potassium

=> $k = \frac{ 80 * 3*10^{-3}}{1}$

$k = 0.240\ kg$

Now from the question potassium-40 is 0.012% of the total potassium so

Amount of potassium-40 present is mathematically represented as

$Z = \frac{0.012}{100} * 0.240$

$Z = 0.0288 *10^{-3}\ kg$

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

$D = \frac{E}{M}$

Substituting values

$D = \frac{1.7622*10^{-13}}{80}$

$D = 2.2*10^{-15} J/kg$

Converting to Sieverts

We have

$D_s = REB * D$

$D_s = 1.2 * 2.2 *10^{-15}$

$D_s = 2.64 *10^{-15}$

So

for half-life ( $1.28 *10^9 \ years$ ) the dose is $2.64 *10^{-15}$

Then for 1 year the dose would be x

=> $x = \frac{2.64 *10^{-15}}{1.28 * 10^9}$

$x = 2.06 *10^{-24}$ per year

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