The drag force on an object with a mass of 1 kg that is falling vertically is FD = C-V2, where c = 0.1 kg/m. What is the terminal velocity? Assume g = 10 m/s2. a. 1 m/s b. 10 m/s c. 100 m/s d. Can't be determined with information provided Explain your reasoning and show your work:
The drag force on an object with a mass of 1 kg that is falling vertically is FD = C-V2, where c = 0.1 kg/m. What is the terminal velocity? Assume g = 10 m/s2. a. 1 m/s b. 10 m/s c. 100 m/s d. Can't be determined with information provided Explain your reasoning and show your work:
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![### Problem Statement:
The drag force on an object with a mass of 1 kg that is falling vertically is given by the equation \( F_D = c \cdot v^2 \), where \( c = 0.1 \, \text{kg/m} \). What is the terminal velocity? Assume \( g = 10 \, \text{m/s}^2 \).
**Options:**
a. 1 m/s
b. 10 m/s
c. 100 m/s
d. Can't be determined with information provided
**Explain your reasoning and show your work:**
---
**Solution:**
At terminal velocity, the drag force equals the gravitational force.
1. **Gravitational Force:**
\[ F_g = m \cdot g = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \]
2. **Drag Force Equation:**
\[ F_D = c \cdot v^2 = 0.1 \, \text{kg/m} \cdot v^2 \]
3. **Equating Forces:**
\[ 10 \, \text{N} = 0.1 \, \text{kg/m} \cdot v^2 \]
4. **Solve for \( v \):**
\[ v^2 = \frac{10 \, \text{N}}{0.1 \, \text{kg/m}} = 100 \, \text{m}^2/\text{s}^2 \]
\[ v = \sqrt{100} = 10 \, \text{m/s} \]
Therefore, the terminal velocity is **10 m/s**. The correct answer is **b. 10 m/s**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14ad990f-dd7d-47e0-a13f-1f780fa26f58%2F7fe8b100-738c-494b-9934-cdf629e0514f%2Fc54jjz2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
The drag force on an object with a mass of 1 kg that is falling vertically is given by the equation \( F_D = c \cdot v^2 \), where \( c = 0.1 \, \text{kg/m} \). What is the terminal velocity? Assume \( g = 10 \, \text{m/s}^2 \).
**Options:**
a. 1 m/s
b. 10 m/s
c. 100 m/s
d. Can't be determined with information provided
**Explain your reasoning and show your work:**
---
**Solution:**
At terminal velocity, the drag force equals the gravitational force.
1. **Gravitational Force:**
\[ F_g = m \cdot g = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \]
2. **Drag Force Equation:**
\[ F_D = c \cdot v^2 = 0.1 \, \text{kg/m} \cdot v^2 \]
3. **Equating Forces:**
\[ 10 \, \text{N} = 0.1 \, \text{kg/m} \cdot v^2 \]
4. **Solve for \( v \):**
\[ v^2 = \frac{10 \, \text{N}}{0.1 \, \text{kg/m}} = 100 \, \text{m}^2/\text{s}^2 \]
\[ v = \sqrt{100} = 10 \, \text{m/s} \]
Therefore, the terminal velocity is **10 m/s**. The correct answer is **b. 10 m/s**.
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