Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
It is a well known fact that the earth rotates around the sun in an inclined axis which is approximately 23 degree. The inclined nature of earth axis causes variation in the solar heat received at any place on the earth surface. The hemisphere facing the sun due to this axial tilt, gets higher sun energy as compared to the opposite side. The hemisphere which faces the sun will experience summer whereas the hemisphere away from sun will experience winter.
In each of the hemisphere the polar areas will receive higher radiation and longer daytime during the summer season. However it has been observed that there is difference in radiation received at different areas of earth surface and radiated. The tropical areas have lower reflectance and thus a large part of incoming solar radiation have been absorbed along the tropics. The poles though have longer daytime during summer and hence greater solar radiation but due to high reflectance radiate more energy. Thus the tropical areas have surplus energy as compared to deficit energy areas of poles. This difference in energy creates a heat imbalance.
This net heat difference between poles and equator gives rise to a global circulation system leading to flow of heat from the net energy excess areas to deficit areas. This circulation takes place through atmosphere as well as oceans and different process of climate viz. evaporation, transpiration, rainfall, wind, convection, oceanic circulations etc work as tools of this system
D is the answer. Hope this helped
Answer:
16-bit wide
Explanation:
In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.
If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:
64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.
In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.
