Answer:
![\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Viscosity%20%5C%20of%20%5C%20glycerine%20%5C%20%28%5Ceta%29%20%3D%2014.382%20%5C%20poise%7D%20)
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine (
)
Explanation:
![\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7Bv%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%28%20%7Br%7D%5E%7B2%7D%20%28%20%5Crho%20-%20%20%5Csigma%29g%29%7D%7B%20%5Ceta%7D%20%7D%7D)
![\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%28%20%7Br%7D%5E%7B2%7D%28%20%5Crho%20-%20%20%5Csigma%29g%20%29%7D%7Bv%7D%20)
Substituting values of r, ρ, σ, v & g in the equation:
![\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%28%20%7B%280.15%29%7D%5E%7B2%7D%20%20%5Ctimes%20%20%287.85%20-%201.25%29%20%5Ctimes%20980.6%29%7D%7B2.25%7D%20)
![\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Cfrac%7B%280.0225%20%5Ctimes%206.6%20%5Ctimes%20980.6%29%7D%7B2.25%7D%20)
![\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Ctimes%20%20%5Cfrac%7B145.6191%7D%7B2.25%7D%20)
![\sf \implies \eta = \frac{2}{9} \times 64.7196](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%20%5Cfrac%7B2%7D%7B9%7D%20%20%5Ctimes%2064.7196)
![\sf \implies \eta = 2 \times 7.191](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%202%20%5Ctimes%207.191)
![\sf \implies \eta = 14.382 \: poise](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%5Ceta%20%3D%20%2014.382%20%5C%3A%20poise)
Answer:
<em>likely to decrease downstream in arid regions and increase downstream in temperate regions</em>
<em></em>
Explanation:
Arid regions are is a region with a severe lack of water, usually to the extent that affect the organisms living in the region. Arid regions are characterized by a very low depth of rainfall per year. Temperate region on the other hand experience more distinct seasonal change and wider temperature change. Temperate regions get a fairly large amount of rainfall per year.
In arid regions, the soil is very dry, and the rate of infiltration and percolation is high relative to the amount of rainfall available. The effect is that more water is infiltrated into the soil as you move downstream, leading to a decrease in the discharge of a stream as you move downstream. Most temperate region have soils that are usually saturated in the peak of the rainfall season, leading to a greater stream discharge as you move downstream.
Answer:
Explanation:
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