Answer:
waxing gibbous is the answer.
Answer:
when a magnet is hanged freely in air it turns in the direction of the north and south while the magnetic north pole faces the south pole of the earth and magnetic south pole faces the north pole if the earth
<span>First law of thermodynamics. This conservation law states that energy cannot be created or destroyed but can be changed from one form to another. In essence, energy is always conserved but can be converted from one form into another. Like when an engine burns fuel, it converts the energy stored in the fuel's chemical bonds into useful mechanical energy and then into heat, or more specifically, the melting ice cubes. Yeast breaks down maltose into glucose to produce alcohol and Co2 in the fermentation process. This is a prime example of the 1st law of thermodynamics. No form of usable energy is really lost; it only changes from one form to another</span>
Answer:
Evaporation
Explanation:
Evaporation is a form of mass tranfer phenomena where by water are moved from the earth surface into the atmosphere as vapours,it is path of the water cycle a decription of the path moved by land water until it turns into rain, humidity,air and temperature are factors that influence evaporation though evaporation can happen at all temperature
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = ![a^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D)
Let the height of the bin be 'h'
Therefore the total area, ![A_{t} = 4ah](https://tex.z-dn.net/?f=A_%7Bt%7D%20%3D%204ah)
The cost is:
C = 2sh
Volume of the box, V =
(1)
Total cost,
(2)
From eqn (1):
![h = \frac{128}{a^{2}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B128%7D%7Ba%5E%7B2%7D%7D)
Using the above value in eqn (1):
![C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}](https://tex.z-dn.net/?f=C%28a%29%20%3D%202a%5E%7B2%7D%20%2B%202a%5Cfrac%7B128%7D%7Ba%5E%7B2%7D%7D%20%3D%202a%5E%7B2%7D%20%2B%20%5Cfrac%7B256%7D%7Ba%7D)
![C(a) = 2a^{2} + \frac{256}{a}](https://tex.z-dn.net/?f=C%28a%29%20%3D%202a%5E%7B2%7D%20%2B%20%5Cfrac%7B256%7D%7Ba%7D)
Differentiating the above eqn w.r.t 'a':
![C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}](https://tex.z-dn.net/?f=C%27%28a%29%20%3D%204a%20-%20%5Cfrac%7B256%7D%7Ba%5E%7B2%7D%7D%20%3D%20%5Cfrac%7B4a%5E%7B3%7D%20-%20256%7D%7Ba%5E%7B2%7D%7D)
For the required solution equating the above eqn to zero:
![\frac{4a^{3} - 256}{a^{2}} = 0](https://tex.z-dn.net/?f=%5Cfrac%7B4a%5E%7B3%7D%20-%20256%7D%7Ba%5E%7B2%7D%7D%20%3D%200)
![\frac{4a^{3} - 256}{a^{2}} = 0](https://tex.z-dn.net/?f=%5Cfrac%7B4a%5E%7B3%7D%20-%20256%7D%7Ba%5E%7B2%7D%7D%20%3D%200)
a = 4
Also
![h = \frac{128}{4^{2}} = 8](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B128%7D%7B4%5E%7B2%7D%7D%20%3D%208)
The path in order to minimize the cost must be a rectangle.