Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, 
= 926 kg/m³
density of the wood, 
= 974 kg/m³
density of water, 
= 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
![F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm](https://tex.z-dn.net/?f=F_%7Bwood%7D%20-%20F_%7Boil%7D%20-%20F_%7Bwater%7D%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.gh%20-%20%5Crho%20_o%20.g%28h-x%29%20-%20%5Crho_w%20.gx%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%20%5Crho%20_o%20%28h-x%29%20-%20%5Crho_w%20.x%20%3D%200%5C%5C%5C%5C%5Crho%20_%7Bwood%7D%20.h%20-%5Crho%20_o%20h%20%2B%20%5Crho%20_o%20x%20-%20%5Crho_w%20.x%20%3D0%5C%5C%5C%5Ch%20%28%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%29%20%3D%20x%28%20%5Crho_w%20-%20%5Crho%20_o%29%5C%5C%5C%5Cx%20%3Dh%5B%5Cfrac%7B%20%5Crho%20_%7Bwood%7D%20%20-%5Crho%20_o%20%7D%7B%5Crho_w%20-%20%5Crho%20_o%7D%20%5D%5C%5C%5C%5Cx%20%3D%203.69%5C%20cm%20%5Ctimes%20%5B%5Cfrac%7B974%20-%20926%7D%7B1000-926%7D%20%5D%5C%5C%5C%5Cx%20%3D%202.39%20%5C%20cm)
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.
Because they are mentally trying to extinguish the negative things in their lives and focus on the positive things
The answer to it is the letter A
Answer:
10.58 ft
Explanation:
Force, F = 1.4 N
Diameter of membrane = 7.4 mm
radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m
Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2
Density, d = 1.03 x 10^3 kg/m^3
Pressure at depth, P = h x d x g
Let h be the depth.
Pressure = force / Area
h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)
h = 3.225 m = 10.58 ft
Thus, the depth of water is 10.58 ft.
