Current passing through a wire as function of time is given as $I(t)=0.02 t+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $t=2 \mathrm{~s}$ is

<p>To find the total charge $ q $ that flows through the wire from $ t = 1 \, \text{s} $ to $ t = 2 \, \text{s} $, we can integrate the current function $ I(t) = 0.02t + 0.01 \, \text{A} $ over this interval.</p> <p>The formula for the charge $ q $ flowing through the wire is given by the integral of the current with respect to time:</p> <p>$ q = \int I(t) \, dt $</p> <p>We integrate the function from $ t = 1 $ to $ t = 2 $:</p> <p>$ q = \int_1^2 (0.02t + 0.01) \, dt $</p> <p>To solve this, first find the antiderivative:</p> <p>$ q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_1^2 $</p> <p>Calculate the expression at the upper and lower bounds:</p> <p>$ = \left[ 0.01t^2 + 0.01t \right]_1^2 $</p> <p>Plug in the values:</p> <p>For $ t = 2 $:</p> <p>$ = 0.01(2)^2 + 0.01(2) = 0.04 + 0.02 = 0.06 $</p> <ol start="2"> <p>For $ t = 1 $:</p> <p>$ = 0.01(1)^2 + 0.01(1) = 0.01 + 0.01 = 0.02 $</p> <p>Subtract the result at $ t=1 $ from the result at $ t=2 $:</p> <p>$ q = 0.06 - 0.02 = 0.04 \, \text{C} $</p> <p>Therefore, the charge that flows through the wire from $ t = 1 \, \text{s} $ to $ t = 2 \, \text{s} $ is $ 0.04 \, \text{C} $.</p>