The attenuated intensity I of a neutron beam normally incident on a certain shie

Question

The attenuated intensity I of a neutron beam normally incident on a certain shielding material is proportional to the intensity of the beam as it passes (the distance x) through the shield, i.e. dI/dx = -mI Where m is the constant of proportionality. If a 25 cm thick shield of this material attenuates the beam by a factor of 1,000, what thickness of the material is required to attenuate the beam by a factor of 5,000? Assume the incident beam has an intensity of I0 when it enters the shield. Given a radioactive particle that encounters a shielding barrier, it is known to decelerate and come to rest within the shielding in 6 microseconds (6 mu s). In this example, presume the distance (d in microns (mu m)) the particle will travel in the time period after hitting the shielding is given by the following equation. Given the bounds t0 = 0 mu s and tmax = 6 mu s d = t3 - 18t2 + 108t What is the maximum distance the particle will travel into the shielding? What is the velocity of the particle at t = 4 mu s? What is the acceleration of the particle at t = 3 mu s? Given the velocity function of a particle to be v(t) = -4e-2t - 2 Find the position function, s(t), that describes where the particle will be at time t in microseconds (mu s). Given that s(0) = 0 microns (mu m), find s(6 mu s).

Explanation / Answer

5)
a) maximum d= d(6) = 6^3 - 18*6^2 + 108*6 = 216 m

b) v(t) = d(d)/dt = 3t^2 - 36*t + 108

so v(4) = 12 m/s

c) a(t) = dv/dt = 6t - 36

a(3) = -18 m/s2

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