A long thin rod lies along the x-axis from the origin to x=L, with L= 0.890 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x2). The value of λ0 is 0.700 kg/m and x is in meters.

1. To calculate the total mass of the rod, we need to integrate the mass density over the length of the rod.

The mass per unit length, λ, is given by λ = λ0 (1 + 1.410x^2), where λ0 = 0.700 kg/m.

To find the total mass, we integrate λ over the length of the rod from x = 0 to x = L:

M = ∫λ dx

M = ∫(λ0 (1 + 1.410x^2)) dx

M = λ0 ∫(1 + 1.410x^2) dx

M = λ0 [x + (1.410/3)x^3] evaluated from x = 0 to x = L

M = λ0 (L + (1.410/3)L^3)

Substituting the given values, L = 0.890 m and λ0 = 0.700 kg/m, we can calculate the total mass.

2. To calculate the x-coordinate of the center of mass of the rod, we need to find the mean position of the mass along the x-axis.

The x-coordinate of the center of mass, x_cm, is given by

x_cm = ∫(λx) dx / M

x_cm = ∫(λ0 (1 + 1.410x^2)x) dx / M

x_cm = λ0/M ∫(x + 1.410x^3) dx

x_cm = λ0/M [(1/2)x^2 + (1.410/4)x^4] evaluated from x = 0 to x = L

x_cm = λ0/M [(1/2)L^2 + (1.410/4)L^4]

Substituting the given values, L = 0.890 m and λ0 = 0.700 kg/m, and the calculated value of M from part 1, we can calculate the x-coordinate of the center of mass.

3. To calculate the moment of inertia of the rod with respect to the y-axis, we need to integrate the squared distance of each small mass element from the y-axis.

The moment of inertia, I, is given by

I = ∫(λx^2) dx

I = ∫(λ0 (1 + 1.410x^2)x^2) dx

I = ∫(λ0 (x^2 + 1.410x^4)) dx

I = λ0 ∫(x^2 + 1.410x^4) dx

I = λ0 [(1/3)x^3 + (1.410/5)x^5] evaluated from x = 0 to x = L

I = λ0 [(1/3)L^3 + (1.410/5)L^5]

Substituting the given values, L = 0.890 m and λ0 = 0.700 kg/m, we can calculate the moment of inertia.