![]() (c) Use C(t) and A(t) found in part (a) to determine the percentage of an initial amount K 0 of K-40 that decays into Ca-40 and the percentage that decays into Ar-40 over a very long period of time. Then use to find and from the first and second equations. (a) From the last equation in the given system of differential equations find K(t) if K ( 0 ) = K 0. By proceeding as in Problem 1 we can solve the foregoing mathematical model. Where λ 1 and λ 2 are positive constants of proportionality. Because the rates at which the amounts C(t) of Ca-40 and A(t) of Ar-40 increase are proportional to the amount K(t) of potassium present, and the rate at which K(t) decays is also proportional to K(t), we obtain the system of linear first-order equations Over time, by emitting beta particles a great percentage of an initial amount K 0 of K-40 decays into the stable isotope calcium-40 (Ca40), whereas by electron capture a smaller percentage of decays into the stable isotope argon-40 (Ar-40). This isotope is also unusual in that it decays by two different nuclear reactions. a, Distribution P ( d) of fragment sizes d for two initial diameters d0 6 mm (red) and d0 12 mm (black). Although potassium occurs naturally in the form of three isotopes, only the isotope potassium-40 (K-40) is radioactive. Figure 4: Distribution of single-drop fragments. Potassium-40 Decay The chemical element potassium is a soft metal that can be found extensively throughout the Earth's crust and oceans. V t = g 4 k ρ k ρ t + r 0 - g 4 k ρ r 0 4 k ρ t + r 0 - 3 This is an exciting site-wide project that demonstrates the cycle and path of rainwater down through the sites landscape: roof, rain garden, creek bed with. ![]() Then, in equation (3), substitute with the value of c, and we get ![]() To get the value of constant c, we must first apply the point of condition ,Ġ = g 4 k ρ k ρ 0 + r 0 + c k ρ 0 + r 0 - 3 0 = g 4 k ρ r 0 + c r 0 - 3 - c r 0 - 3 = g k ρ 4 r 0 c = - g k ρ r 0 r 0 3 4 = - g 4 k ρ r 0 4 V t = g 4 k ρ k ρ t + r 0 + c k ρ t + r 0 - 3 ∫ d k p t + r 0 3 v = g ∫ k p t + r 0 3 d t k ρ t + r 0 3 v = g × 1 4 k ρ k ρ t + r 0 4 + c Then, if we reverse one step, we can arrive at K ρ t + r 0 3 d v d t + k ρ t + r 0 3 3 k ρ k ρ t + r 0 v = g k ρ t + r 0 3 k ρ t + r 0 3 d v d t + 3 k ρ k ρ t + r 0 3 = g k ρ t + r 0 3 Multiply both sides of the first order differential equation stated in (1) by this integrating factor, and we get This differential equation has the form of a first-order linear D.E.įor this D.E, we need to find an integrating factor F.į = e ∫ p ( t ) d t = e ∫ 3 k ρ k ρ t + r 0 d t = e 3 ∫ k ρ k ρ t + r 0 d t = e 3 l n k ρ t + r 0 d t = k ρ t + r 0 3 To study the shape size relation, an observation domain is introduced. dr, and specific differential propagation phase K. ![]() As the first order differential equation, we have a raindrop that falls in a spherical shape and evaporates while maintaining its shape with no air resistance. This paper presents a technique by which it is possible to retrieve the drop shape-size relation that governs the polarimetric radar observations of reflectivity, Z. ![]()
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