**Homework 2:**A 35 cm diameter solid sphere $(\rho = 4500 \, \text{kg/m}^3)$ is suspended by a cable as shown below. Half of the sphere is in one fluid $(\rho = 1200 \, \text{kg/m}^3)$ and the other half is in another $(\rho = 1500 \, \text{kg/m}^3)$. What is the tension in the cable?---**Diagram Explanation:**The diagram illustrates a solid sphere partially submerged in two different fluids. The sphere is held by a cable from above. - The diameter of the sphere is indicated as 35 cm.- The top half of the sphere is submerged in a fluid with a density of 1200 kg/m$^3$.- The bottom half of the sphere is submerged in a fluid with a density of 1500 kg/m$^3$.- The total height of submersion is labeled as 120 cm.This setup is used to determine the tension in the cable supporting the sphere.

Answered: HW 2 A 35 cm diam Suspended by a

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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**Homework 2:**

A 35 cm diameter solid sphere $(\rho = 4500 \, \text{kg/m}^3)$ is suspended by a cable as shown below. Half of the sphere is in one fluid $(\rho = 1200 \, \text{kg/m}^3)$ and the other half is in another $(\rho = 1500 \, \text{kg/m}^3)$.

What is the tension in the cable?

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**Diagram Explanation:**

The diagram illustrates a solid sphere partially submerged in two different fluids. The sphere is held by a cable from above.

- The diameter of the sphere is indicated as 35 cm.
- The top half of the sphere is submerged in a fluid with a density of 1200 kg/m$^3$.
- The bottom half of the sphere is submerged in a fluid with a density of 1500 kg/m$^3$.
- The total height of submersion is labeled as 120 cm.

This setup is used to determine the tension in the cable supporting the sphere.

Expert Solution

Step 1: Write the given data with suitable variables-

$D equals 35 space c m rho subscript s equals 4500 space kg divided by straight m cubed rho subscript 1 equals 1200 space kg divided by straight m cubed straight rho subscript 2 equals 1500 space kg divided by straight m cubed V subscript 1 equals Volume space of space body space immersed space in space liquid space 1 V subscript 2 equals Volume space of space body space immersed space in space liquid space 2 V subscript s equals space 4 over 3 pi R cubed equals space 4 over 3 pi open parentheses fraction numerator 0.35 over denominator 2 end fraction close parentheses cubed equals 0.022449 space straight m cubed$

The free-body diagram is as follows-

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