What force is needed to pull one microscope slide over another at a speed of 1.00 cm/s, if there is a 0.500-mm-thick layer of 20 degrees C water between them and the contact area is 8.00 cm^2?

To find the force required to pull one microscope slide over another at a speed of 1.00 cm/s with a 0.500-mm-thick layer of water between them, we need to use the concept of viscous drag force. The formula for viscous drag force \( F \) in this context is given by:

\[
F = \eta \frac{A v}{d}
\]

where:
- \( \eta \) is the dynamic viscosity of the fluid (water in this case),
- \( A \) is the contact area,
- \( v \) is the velocity,
- \( d \) is the thickness of the fluid layer.

Let's break this down:

1. **Dynamic viscosity of water (\( \eta \))**: At 20°C, the viscosity of water is approximately \( 1.002 \times 10^{-3} \, \text{Pa} \cdot \text{s} \).

2. **Contact area (\( A \))**: Given as 8.00 cm², we convert this to m²:
   \[
   A = 8.00 \, \text{cm}^2 = 8.00 \times 10^{-4} \, \text{m}^2
   \]

3. **Velocity (\( v \))**: Given as 1.00 cm/s, we convert this to m/s:
   \[
   v = 1.00 \, \text{cm/s} = 1.00 \times 10^{-2} \, \text{m/s}
   \]

4. **Thickness of the fluid layer (\( d \))**: Given as 0.500 mm, we convert this to meters:
   \[
   d = 0.500 \, \text{mm} = 0.500 \times 10^{-3} \, \text{m} = 5.00 \times 10^{-4} \, \text{m}
   \]

Now, substituting these values into the formula:

\[
F = (1.002 \times 10^{-3} \, \text{Pa} \cdot \text{s}) \frac{(8.00 \times 10^{-4} \, \text{m}^2)(1.00 \times 10^{-2} \, \text{m/s})}{5.00 \times 10^{-4} \, \text{m}}
\]

Simplifying:

\[
F = \frac{1.002 \times 10^{-3} \times 8.00 \times 10^{-4} \times 1.00 \times 10^{-2}}{5.00 \times 10^{-4}} \, \text{N}
\]

\[
F = \frac{1.002 \times 8.00 \times 10^{-9}}{5.00} \, \text{N}
\]

\[
F = \frac{8.016 \times 10^{-9}}{5.00 \times 10^{-4}} \, \text{N}
\]

\[
F = 1.6032 \times 10^{-5} \, \text{N}
\]

\[
F \approx 1.60 \times 10^{-5} \, \text{N}
\]

Therefore, the force required is approximately **\( 1.60 \times 10^{-5} \, \text{N} \)**.

Would you like more details or have any questions? Here are some related questions:

1. How does the viscosity of water change with temperature?
2. What is the relationship between force and the thickness of the fluid layer?
3. How would the required force change if the contact area were doubled?
4. What if the fluid between the slides were oil instead of water?
5. How does the velocity affect the viscous drag force?

**Tip:** When dealing with fluid dynamics problems, always ensure that units are consistent to avoid errors in your calculations.

Learn how to calculate the force required to pull one microscope slide over another at 1.00 cm/s, with a 0.500-mm-thick layer of 20°C water and an 8.00 cm² contact area. This problem involves fluid dynamics and the application of the viscous drag force formula. Suitable for advanced high school students studying physics or engineering.

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