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3 years ago

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A patient is administered 20 mg of iodine-131. How much of this isotope will remain in the body after 40 days if the half-life f

or iodine-131 is 8 days?

1 answer:

igor_vitrenko [27]3 years ago

4 0

The formula for half-life is:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where A is the amount of iodine-131 initially and after 40 days, t is time, h is half-life of the isotope. Let's plug in our values to the equation:

$A_{final}=20(\frac{1}{2})^{\frac{40}{8}=0.625g$

Therefore, the patient has 0.625 grams of iodine-131 after 40 days.

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The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

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T = $\frac{2 \pi}{4 \pi}$

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Answer:

The magnitude of the electric field is 5.75 N/C towards positive x- axis.

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Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})$

Put the value into the formula

$E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})$

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The direction is toward positive x- axis.

Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.

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Answer:

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