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:
The correct option is c.
Explanation:
Metabolism is a sum of anabolic and catabolic reactions. The body's inability to produce/synthesize enough insulin is the cause of type II diabetes. Generally, metabolism is the process in which most compounds (proteins, carbohydrates and lipids) are produced (anabolism) or broken down (catabolism) in the body. Insulin is a protein that can be produced in less amount due to metabolic disorder in the body.
Maria's disease means she already has an exponentially high amount of blood sugar against the required insulin to balance it out, hence the disease already slowed down her rate of metabolism (catabolism) of blood sugar EXCEPT she decides to increase of metabolism by medication and exercise.
Answer:
The angular velocity at the beginning of the interval is
.
Explanation:
Given that,
Angular acceleration 
Angular displacement 
Angular velocity 
We need to calculate the angular velocity at the beginning
Using formula of angular velocity


Where,
= angular acceleration
= angular velocity
Put the value into the formula



Hence, The angular velocity at the beginning of the interval is
.
Answer:
The wavelength of the wave is 20 m.
Explanation:
Given that,
Amplitude = 10 cm
Radial frequency 
Bulk modulus = 40 MPa
Density = 1000 kg/m³
We need to calculate the velocity of the wave in the medium
Using formula of velocity

Put the value into the formula


We need to calculate the wavelength
Using formula of wavelength


Put the value into the formula


Hence, The wavelength of the wave is 20 m.
density of water = 
velocity of flow = 
radius of pipe = 
Height of second floor = 
Now we can use here Bernuoli's Equation to find the speed of water flow at second floor



Now in order to find the radius of pipe we can use equation of continuity



So radius of pipe at second floor is 0.034 meter