Question

help pls!!!!!

Consider a tank filled with a liquid of constant density ρ up to an initial height ho in Earth. If the tank is moving with a constant linear acceleration ao, it is known from Fluidstatics that the isobars (lines with constant pressure) and therefore the free surface become a straight Line whose slope is a linear function of ao. If a0-0 then the surface becomes flat. On the other hand, if the tank is subject to rigid body rotation characterized by Ω, then the isobars are paraboloids of revolution being its curvature proportional to Ω Determine the analytical expressions of the isobars s (, yc for a liquid inside a tank moving with linear acceleration a >0 along the 2 axis and rotating with a non-zero constant angular speed Ω about the vertical axis Note: it will be convenient to analyze the problem taking a non-inertial frame of reference moving (and rotating) with the tank. Select one: b. c. I would rather not answer this question 2g z,(x,y) = 2g

Answer 1

Given data: - Contact Density = e Initial height = mu_0 Linear Acceleration = a_0 Now, from Euler eq: - [alpha p = eh^2 w^2/M dM-egd^2] Now, considering p section: - integral^p2_p1 dp = ew^2 integral^H_2_H-1 M.DM-eg integral^22_21 dx p_2-p_1 = ew^2/2(M^2_2-M^2_1)-eg(z_2-z_1) p alpha M^2, p alpha w^2 i.e p alpha ohm^2 Now, At location we have atmospheric pressure: - 0 = ew^2/2 (M^2_2-M^2_1)-eg(z_2-z_1) g(z_2-z_1) = w^2/2(M^2_2-M^2_1) z_2-z_1 = w^2/2g (M^2_2-M^2_1) \r\n Now, comparing the equations from the expression: - i.e z-1(x, y) = ohm^2/2g[(x + y a_0/ohm^2)^2 - a^2_0/ohm^4] where M_2 = x + y - a-0/ohm^2 M-1 = a-0/ohm^2 When, the initial volume of liquid is equal to final volume of liquid that w-0 spilling out of water then x = y The answer is z_1(x, y) = ohm^2/2g [(x + y-a_0/ohm^2)^2-a^2_0/ohm^4 Hence, we need to compare the equation with the equation so obtain we get the desired equation as answer where ohm = w