**Problem Statement:**6) A tower used in a wind turbine system had a stiffness of \(790 \, \text{N/m}\) after displacing \(10 \, \text{cm}\). Find the weight (in grams) that was connected to this tower. Assume \(g_L = \frac{10.174 \, \text{m}}{\text{s}^2}\). **Explanation:**1. **Stiffness and Displacement:** - Stiffness (\(k\)) is the resistance of an elastic body to deformation, given as \(790 \, \text{N/m}\). - The displacement (\(x\)) is \(10 \, \text{cm}\) which should be converted to meters: \(x = 0.1 \, \text{m}\).2. **Force Calculation with Hooke’s Law:** - Hooke’s Law: \(F = k \times x\). - Calculate the force (\(F\)) exerted: \(F = 790 \, \text{N/m} \times 0.1 \, \text{m} = 79 \, \text{N}\).3. **Weight and Mass Relationship:** - Weight (\(W\)) is the force due to gravity: \(W = m \times g\). - Rearrange to solve for mass (\(m\)): \(m = \frac{W}{g_L}\). 4. **Substitute Values to Find Mass:** - Given \(g_L = \frac{10.174 \, \text{m}}{\text{s}^2}\). - Mass (\(m\)) in kilograms: \(m = \frac{79 \, \text{N}}{10.174 \, \text{m/s}^2}\).5. **Convert Mass to Grams:** - Convert from kilograms to grams by multiplying by 1000.**Note:**- This problem involves understanding the concepts of mechanical stiffness, force, displacement, and weight. The conversion of weight to mass using a specific gravitational value is essential for engineering applications.

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Answered: 6) A tower used in a wind turbine…

6) A tower used in a wind turbine system had a stiffness of 790N/m after displacing 10cm. Find the weight(in gram) that was connected to this tower. Assume gi= 10.174 m s2 +.

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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Question

**Problem Statement:**

6) A tower used in a wind turbine system had a stiffness of \(790 \, \text{N/m}\) after displacing \(10 \, \text{cm}\). Find the weight (in grams) that was connected to this tower. Assume \(g_L = \frac{10.174 \, \text{m}}{\text{s}^2}\).

**Explanation:**

1. **Stiffness and Displacement:**
- Stiffness (\(k\)) is the resistance of an elastic body to deformation, given as \(790 \, \text{N/m}\).
- The displacement (\(x\)) is \(10 \, \text{cm}\) which should be converted to meters: \(x = 0.1 \, \text{m}\).

2. **Force Calculation with Hooke’s Law:**
- Hooke’s Law: \(F = k \times x\).
- Calculate the force (\(F\)) exerted: \(F = 790 \, \text{N/m} \times 0.1 \, \text{m} = 79 \, \text{N}\).

3. **Weight and Mass Relationship:**
- Weight (\(W\)) is the force due to gravity: \(W = m \times g\).
- Rearrange to solve for mass (\(m\)): \(m = \frac{W}{g_L}\).

4. **Substitute Values to Find Mass:**
- Given \(g_L = \frac{10.174 \, \text{m}}{\text{s}^2}\).
- Mass (\(m\)) in kilograms: \(m = \frac{79 \, \text{N}}{10.174 \, \text{m/s}^2}\).

5. **Convert Mass to Grams:**
- Convert from kilograms to grams by multiplying by 1000.

**Note:**
- This problem involves understanding the concepts of mechanical stiffness, force, displacement, and weight. The conversion of weight to mass using a specific gravitational value is essential for engineering applications.

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