Answer:

& 
Explanation:
Given:
- interior temperature of box,

- height of the walls of box,

- thickness of each layer of bi-layered plywood,

- thermal conductivity of plywood,

- thickness of sandwiched Styrofoam,

- thermal conductivity of Styrofoam,

- exterior temperature,

<u>From the Fourier's law of conduction:</u>

....................................(1)
<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>




.....................(2)
Putting the value from (2) into (1):


is the heat per unit area of the wall.
The heat flux remains constant because the area is constant.
<u>For plywood-Styrofoam interface from inside:</u>



&<u>For Styrofoam-plywood interface from inside:</u>



Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2