Answer:
q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s
Explanation:
For laminar flow between two parallel horizontal plates, the volumetric flow per metre of width is given as
q = (2h³/3μ) (ΔP/L)
h = hydraulic depth = 4mm/2 = 2mm = 0.002 m
μ = viscosity of oil (SAE 30) at 15.6°C = 0.38 Pa.s
(ΔP/L) = 26 KPa/m = 26000 Pa/m
q = (2h³/3μ) (ΔP/L)
q = (26000) × (2(0.002³)/(3×0.38))
q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s
Answer:
D. a windstorm scratches the surface of a rock with sand.
Answer:
The Answer is gonna be B.
Air warms and rises.
In order to get the magnitude of the electric field at the location of the test charge, you have to use the formula: E=Fe/q.
Given E as electric field strength; Fe as electric force = 0.751 N and q as charge = 5*10-5 C. Therefore,
E= 0.751 N / 5 * 10-5 CE=15020 N C-1 or V m-1 Electric field at the location.