To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as 
, the relation with the final volume as
Initial temperature = 
Let T be the temperature after expanding by the formula of volume expansion
we have,
Where 
is the volume coefficient of copper 
Therefore the temperature is 53.06°C
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)
Distance traveled is
d = 100 km = 10⁵ m
Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s
Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s
The difference in time is
302.12 - 19.49 = 282.63 s
Answer: 283 s (nearest second)
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.
