You do it by using what you are given. Unfortunately, you haven't mentionedwhat that is, so we can't be any more specific.
3y + x = k where k is some constant which can only be determined if a point on it is known. There is no such point given.
Boltzman constant
limx→∞(k)=k, the limit of a constant k is equal to the constant k. Therefore, the limx→∞(6)=6.
what is the constant variation of k if x=.5 and y=1
== == ---- ----
Gold never decays by alpha emission, it either decays by -beta, +beta, K capture, or gamma emission depending on isotope.Natural gold is isotopically pure gold-197, which is stable.
First order rate constant k is described in V=k[EA] while second order rate constant is given as V=k[E][A]. For reactions that do not have true order, k is the apparent rate constant.
Npn decays to Pan-4 and alpha. Only isotopes 234, 235, and 237 of neptunium can undergo alpha decay, the others decay by beta-, beta+, K capture, and/or gamma decay. So the only products of neptunium alpha decay can be protactinium isotopes 230, 231, or 233.
The force of repulsion between the alpha particle and the gold nucleus can be calculated using Coulomb's law, given by F = k * (q1 * q2) / r^2, where k is the Coulomb constant, q1 and q2 are the charges of the particles, and r is the distance between them. Given the charges of an alpha particle and a gold nucleus, and the distance of 1pm, the force of repulsion can be calculated to be extremely large due to the proximity of the particles and the high charges involved.
The alphabet!The corret answer is alpha,beta,chi,delta,epsilon, gama,iota,kapa,lamdamunuomeaga pi sigma theta or tau and zeda
The alphabet!The corret answer is alpha,beta,chi,delta,epsilon, gama,iota,kapa,lamdamunuomeaga pi sigma theta or tau and zeda
The expression for the force constant (k) in Hooke's Law is given by the equation F = kx, where F is the force applied, k is the force constant, and x is the displacement from equilibrium. The force constant is a measure of the stiffness of a spring or a bond.
Think you've got this backwards. The exponential probability distribution is a gamma probability distribution only when the first parameter, k is set to 1. Consistent with the link below, if random variable X is distributed gamma(k,theta), then for gamma(1, theta), the random variable is distributed exponentially. The gamma function in the denominator is equal to 1 when k=1. The denominator will reduce to theta when k = 1. The first term will be X0 = 1. using t to represent theta, we have f(x,t) = 1/t*exp(-x/t) or we can substitute L = 1/t, and write an equivalent function: f(x;L) = L*exp(-L*x) for x > 0 See: http://en.wikipedia.org/wiki/Gamma_distribution [edit] To the untrained eye the question might seem backwards after a quick google search, yet qouting wikipedia lacks deeper insight in to the question: What the question is referring to is a class of functions that factor into the following form: f(y;theta) = s(y)t(theta)exp[a(y)b(theta)] = exp[a(y)b(theta) + c(theta) + d(y)] where a(y), d(y) are functions only reliant on y and where b(theta) and c(theta) are answers only reliant on theta, an unkown parameter. if a(y) = y, the distribution is said to be in "canonical form" and b(theta) is often called the "natural parameter" So taking the gamma density function, where alpha is a known shape parameter and the parameter of interest is beta, the scale parameter. The density function follows as: f(y;beta) = {(beta^alpha)*[y^(alpha - 1)]*exp[-y*beta]}/gamma(alpha) where gamma(alpha) is defined as (alpha - 1)! Hence the gamma-density can be factored as follows: f(y;beta) = {(beta^alpha)*[y^(alpha - 1)]*exp[-y*beta]}/gamma(alpha) =exp[alpha*log(beta) + (alpha-1)*log(y) - y*beta - log[gamma(alpha)] from the above expression, the canonical form follows if: a(y) = y b(theta) = -beta c(theta) = alpha*log(beta) d(y) = (alpha - 1)*log(y) - log[gamma(alpha)] which is sufficient to prove that gamma distributions are part of the exponential family.
You do it by using what you are given. Unfortunately, you haven't mentionedwhat that is, so we can't be any more specific.
3y + x = k where k is some constant which can only be determined if a point on it is known. There is no such point given.
Boltzman constant