A wet substance in the open air loses its moi | vedclass

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(A) Let $y$ be the amount of moisture at time $t$.The rate of change of moisture is proportional to the moisture content:$\frac{dy}{dt} = -ky$ (where

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$\frac{2 \log 10}{\log 2}$

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(A) Let $y$ be the amount of moisture at time $t$.The rate of change of moisture is proportional to the moisture content:$\frac{dy}{dt} = -ky$ (where $k > 0$ is a constant).Separating variables and integrating:$\int \frac{dy}{y} = -\int k dt \Rightarrow \ln y = -kt + C$.At $t = 0$,let the initial moisture be $y_0 = 1$ (representing $100 \%$).Then $\ln(1) = -k(0) + C \Rightarrow C = 0$.So,$\ln y = -kt$.Given that at $t = 1$ hour,the sheet loses half its moisture,so $y = 0.5$ (or $1/2$):$\ln(0.5) = -k(1) \Rightarrow k = -\ln(0.5) = \ln(2)$.Now,we need to find $t$ when $99 \%$ of the moisture is lost,meaning $1 \%$ remains:$y = 0.01 = \frac{1}{100} = 10^{-2}$.Substituting into the equation $\ln y = -kt$:$\ln(10^{-2}) = -(\ln 2)t$.$-2 \ln(10) = -(\ln 2)t$.$t = \frac{2 \ln 10}{\ln 2} = \frac{2 \log 10}{\log 2}$.

$A$ wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the open air loses half its moisture during the first hour,then the time $t$,in which $99 \%$ of the moisture will be lost,is

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