Calculating Cs-137 Flux and Fall-Off at Various Heights in a 10 km Field

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To calculate the center line flux of Cs-137 at various heights above a 10 km diameter field, one must assume uniform surface deposition of 100 Curie’s. The flux can be determined by integrating over the area, taking into account the height above the ground. As the height increases, the flux will decrease, following an inverse square law pattern. Specific calculations for heights of 1 meter, 10 meters, 100 meters, and 1000 meters reveal how the flux diminishes with distance. Understanding this relationship is crucial for accurate assessments of radiation exposure.
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A Cs-137 cloud deposits 100 Currie’s on farmer John’s field which is 10 km in diameter. What is the center line flux at one meter above the ground? Recalculate flux if you are 10 meters, 100 meters, and 1000 meters above the ground? How does the flux fall off as a function of distance?

One of my homework problems I'm not sure how to do. Can anybody help me? or start me off in the right direction
 
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Hi there,

You have to suppose a uniform surface deposition. From there, with a scheme of the situation, you should see that you need to simply integrate.

Cheers
 
Thread 'Have I solved this structural engineering equation correctly?'

Thread 'Have I solved this structural engineering equation correctly?'

Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
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