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A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plate located at a distance of 0.01 cm from it, the fluid separating them being water with viscosity of 0.001 Ns/m2. The power required to maintain velocity will be?
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A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plat...
Calculation of Power Required to Maintain Velocity

Given parameters:
- Area of the plate (A) = 0.1 m^2
- Velocity of the plate (v) = 30 cm/s = 0.3 m/s
- Distance between the plates (d) = 0.01 cm = 0.0001 m
- Viscosity of the fluid (μ) = 0.001 Ns/m^2

Formula used:
- Shear stress (τ) = μ(dv/dy)
- Power (P) = τAv

Explanation:
- The shear stress is the force per unit area that acts tangentially to the surface of the plate due to viscosity.
- The shear stress is proportional to the velocity gradient (dv/dy) and the viscosity (μ).
- As the distance between the plates is very small, the velocity gradient can be assumed to be constant.
- The power required to maintain the velocity is the product of shear stress, area, and velocity.

Calculation steps:
- Velocity gradient (dv/dy) = v/d = 0.3/0.0001 = 3000 s^-1
- Shear stress (τ) = μ(dv/dy) = 0.001 x 3000 = 3 N/m^2
- Power (P) = τAv = 3 x 0.1 x 0.3 = 0.09 W

Therefore, the power required to maintain the velocity is 0.09 W.
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A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plate located at a distance of 0.01 cm from it, the fluid separating them being water with viscosity of 0.001 Ns/m2. The power required to maintain velocity will be?
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