This is a mobile phone mast. Underground there are fibre cables to the mast. The material that
these cables are made out of is a flexible and because they are
transparent the pulses of
_that travel through them can do so
very quickly. These masts are always positioned in places that do not have a
lot of buildings or trees that will cause
The
the
mast the wider the area that it will cover. These days mobile masts can deliver
2G, 3G, 4G or
When you make a phone call your phone converts your voice to
waves that travel to the nearest mast. The mast then sees if the call is to a
landline or another mobile phone and diverts the call to where it needs to go.
The mobile phone operator breaks the UK into sections or 'cells' and each cell
contains a
Each cell covers its small area. This is why mobile
phones are called 'cell phones' in other countries such as
Look at
the image below so see how areas are split into cells. The more densely
populated the area the more masts there will be therefore you will see more
masts in
areas that you would in
areas.

Norder to receive financial aid at his vocational school, mario must fill out the fafsa. the fafsa is a form that must be completed to determine . in order to complete a fafsa, you must submit . the fafsa can students obtain

When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2ă—, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5ă— slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?

Me and category do i put them in because this is science