Suppose a car is molded as a cylinder moving on earth at a speed v. If A is the area of cross section of the car and `phi` is the density of air then

Options

A. Power loss due to air resistance is `(A rho v^(3))/(2)`
B. Power loss due to air resistance is A `rho v^(3)`
C. drag force is `( A rho v^(2))/(2)`
D. charge force is `A rho v^(2)`

Answer

To solve the problem regarding the power loss due to air resistance for a car modeled as a cylinder moving at speed \( v \), we will follow these steps:

### Step 1: Understand the Drag Force Formula
The drag force \( F_D \) acting on an object moving through a fluid (in this case, air) can be expressed using the formula:
\[ F_D = \frac{1}{2} C_D \rho A v^2 \]
where:
- \( C_D \) is the drag coefficient (approximately 1 for a cylindrical shape),
- \( \rho \) is the density of the fluid (air),
- \( A \) is the cross-sectional area of the object,
- \( v \) is the velocity of the object relative to the fluid.

### Step 2: Substitute the Values
Since the car is modeled as a cylinder and we are given that the drag coefficient \( C_D \) is approximately 1, we can simplify the drag force to:
\[ F_D = \frac{1}{2} \rho A v^2 \]

### Step 3: Calculate the Power Loss Due to Drag
The power loss \( P \) due to air resistance can be calculated using the formula:
\[ P = F_D \cdot v \]
Substituting the expression for \( F_D \) from Step 2:
\[ P = \left(\frac{1}{2} \rho A v^2\right) \cdot v \]
This simplifies to:
\[ P = \frac{1}{2} \rho A v^3 \]

### Step 4: Conclusion
The power loss due to air resistance for the car moving at speed \( v \) is given by:
\[ P = \frac{1}{2} \rho A v^3 \]

### Summary
Thus, the power loss due to air resistance as the car moves through the air can be expressed in terms of the air density \( \rho \), the cross-sectional area \( A \), and the speed \( v \).

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