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2 years ago

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An electron of mass 9.1110-31kg has an initial speed of 3.00105m/s. It travels in a straight line and its speed increases to 7

.00*105m/s in a distance of 5.00cm. Assuming its acceleration is constant, (a) determine the magnitude of the force exerted on the electron (b) Compare this force (F) with the weight of the electron (Fg), which we ignored

1 answer:

s2008m [1.1K]2 years ago

3 0

Answer:

a) F = 36.4 10⁻¹⁹ N b) F /W = 4.08 10¹¹

Explanation:

a) To find the force we will use Newton's second law and to find the acceleration we will use the kinematics equations

$v_{f}$ ² = v₀² + 2 a x

a = ( $v_{f}$ ²-v₀²) / 2x

a = ((7 10⁵)²2 - (3 10⁵)²) / 2 0.05

a = 4.00 10¹² m /s²

With Newton's second law we calculate the force

F = m a

F = 9.11 10⁻³¹ 4 10¹²

F = 36.4 10⁻¹⁹ N

b) calculate the weight of the electron

W = mg

W = 9.11 10⁻³¹ 9.8

W = 89.3 10⁻³¹ N

The comparison is made by dividing the two magnitudes

F / W = 36.4 10⁻¹⁹ / 89.3 10⁻³¹

F /W = 4.08 10¹¹

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gregori [183]

Complete Question

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

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

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