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3 years ago

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An acceptable radiograph was taken using 40 mAs at 80 kVp at a distance of 60 inches. A second radiograph is requested at 40 inc

hes. What mAs should be used to produce this radiograph with a 40-inch distance

1 answer:

oee [108]3 years ago

4 0

Answer:

17.8 mAs

Explanation:

The exposure maintenance formula shows that as SID increases, intensity decreases, causing a decrease in film exposure and density. The mAs is directly proportional to the square of the distance. That is as mAs increases, distance increases, and vice versa, in order to maintain image receptor exposure. It is given by:

$\frac{mAs_1}{mAs_2}=\frac{D_1^2}{D_2^2}$

Given that mAs₁ = original mAs = 40 mAs, D₁ original distance = 60 in, D₂ = new distance = 40 in, mAs₂ = new mAs

$\frac{mAs_1}{mAs_2}=\frac{D_1^2}{D_2^2}\\\\mAs_2=\frac{mAs_1*D_2^2}{D_1^2}=\frac{40*40^2}{60^2}=17.8\\\\mAs_2=17.8\ mAs$

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$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

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$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

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