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Nataly_w [17]
3 years ago
7

At a distance of 41 ft , an ionizing radiation source delivers 5.0 rem of radiation. How close could you get to the source and s

till have no biological effects? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
mash [69]3 years ago
6 0

Answer:

at distance 18.33 ft no biological effects

Explanation:

given data

distance = 41 ft

radiation = 5 rem

to find out

how close get so no biological effects

solution

we know here that intensity of radiation R is inversely proportional to (radiation)²

so here equation will be

\frac{I1}{I2}  = (\frac{R2}{R1})^{2}   .............1

here I is intensity

so here I1 = 25

because for  0 to 25 there is  no detectable effects and for 25 to 100 it will . temporary decrease so

we take 25 because dose is less than 25 so we take that highest value

so

\frac{I1}{I2}  = (\frac{R2}{R1})^{2}

\frac{25}{5}  = (\frac{41}{R1})^{2}

R1 = 18.33 ft

so at distance 18.33 ft no biological effects

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If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinit
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W = 0×(kq2L)

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We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r

If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them

WABC = kq2/(rAB + rAC + rBC)

So, to assemble four charges A, B, C, & D....

WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)

 

Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative

rAB = L

rAC = L√2

rAD = L (-)

rBC = L

rBD = L√2 (-)

rCD = L (-)

⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)

⇒ ∴ W = 0 × (kq2/L)

This way, working through each option...  

(a)

The positive charges are equidistant from each other at a distance of L.

rAB = L

rAC = L

rAD = ½L⋅sin(60) (-)

rBC = L

rBD = ½L⋅sin(60) (-)

rCD = ½L⋅sin(60) (-)

Wa = kq2/(3L - (3/2)L⋅(0.866))

⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)

(b)

rAB = L

rAC = 2L

rAD = 3L (-)

rBC = L

rBD = 2L (-)

rCD = L (-)

Wb = kq2/(4L - 6L)

⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)

(c)

The factor doesn't matter, so Wc = 0 × (kq2/L)

In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.  

This way, charge configuration (a) requires the least amount of work.

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