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A large open top container of negligible...

A large open top container of negligible mass and uniform cross sectional area `A` a has a small hole of cross sectional area `A//100` in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density `rho` and mass `M_(0)`. Assuming that the liquid starts flowing out horizontally through the hole at `t=0`, calculate a the acceleration of the container and its velocity when `75%` of the liquid has drained out.

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To solve the problem, we need to find the acceleration of the container and its velocity when 75% of the liquid has drained out. Here’s a step-by-step solution: ### Step 1: Understand the System We have a large open-top container with a small hole at the bottom. The container has a cross-sectional area \( A \) and the hole has a cross-sectional area \( \frac{A}{100} \). The liquid inside has a density \( \rho \) and an initial mass \( M_0 \). ### Step 2: Determine the Mass of Liquid Remaining When 75% of the liquid has drained out, 25% remains. Thus, the mass of the liquid left in the container is: \[ ...
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