Home / Expert Answers / Calculus / the-truncated-conical-container-shown-to-the-right-is-full-of-a-beverage-that-weighs-0-55-math-pa618

(Solved): The truncated conical container shown to the right is full of a beverage that weighs \( 0.55 \math ...



The truncated conical container
shown to the right is full of a
beverage that weighs \( 0.55 \mathrm{oz} / \mathrm{in}^{3} \)

The truncated conical container shown to the right is full of a beverage that weighs \( 0.55 \mathrm{oz} / \mathrm{in}^{3} \). The container is 9 in. deep, \( 2.8 \) in. across at the base, and \( 3.6 \) in. across at the top. A straw sticks up 4 in. above the top. How much work does it take to suck up the beverage through the straw (neglecting friction)? Let \( y=0 \) correspond to the bottom of the container. Set up the integral that gives the work required, in in.-oz, to suck up the beverage through the straw (neglecting friction). \[ W=\int d y \]


We have an Answer from Expert

View Expert Answer

Expert Answer


The value of density of truncated container is ?=0.55 oz/in 3 From the schemat
We have an Answer from Expert

Buy This Answer $5

Place Order

We Provide Services Across The Globe