A thin rod extends from x = 0 to x = 15.0 cm. It has a...

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A thin rod extends from x = 0 to x = 15.0 cm. It has a cross-sectional area A = 7.50 cm, and Its density increases uniformly in the positive x-direction from 2.00 g/cm at one endpoint to 19.5 g/cm at the other. (a) The density as a function of distance for the rod is given by p= B + Cx, where B and C are constants. What are the values of B (in g/cm) and C (in g/cm^)? B = g/cm3 9/cm4 C = (b) Finding the total mass of the rod requires integrating the density function over the entire length of the rod. The integral is written as follows. m = famp dv - 1 PA dx = [15.0 cm (B+Cx)(7.50 cm2) dx Jallmaterial Jall x What is the total mass of the rod (in kg)? kg Need Help? Submit Answer Read It A thin rod extends from x = 0 to x = 15.0 cm. It has a cross-sectional area A = 7.50 cm, and Its density increases uniformly in the positive x-direction from 2.00 g/cm at one endpoint to 19.5 g/cm at the other. (a) The density as a function of distance for the rod is given by p= B + Cx, where B and C are constants. What are the values of B (in g/cm) and C (in g/cm^)? B = g/cm3 9/cm4 C = (b) Finding the total mass of the rod requires integrating the density function over the entire length of the rod. The integral is written as follows. m = famp dv - 1 PA dx = [15.0 cm (B+Cx)(7.50 cm2) dx Jallmaterial Jall x What is the total mass of the rod (in kg)? kg Need Help? Submit Answer Read It