what do Mhz  or  Ghz mean?
356325
megahertz and gigahertz. More computer space available for storage.
605350
The terms that you are using are from the IEEE (and older IRE) band designations that have been around for at least 40 years.\nKu stands for (K)urz(German for short) (u)nder and runs from 12 to 18 GHz.\nKa stands for (K)urz (a)bove and runs from 18 to 40 GHz\nC stands for "compromise" and runs from 4 to 6 GHz\nAll of the IEEE designations are: I,G,P,L,S,C,X,Ku,K,Ka,V and W\nAs you can see these designations are a way to name the microwave bands.\nNATO and the European Union have a different designation and they are: A,B,C,D,E,F,G,H,I,J,K,L and M.\nThe only correlation (in terms of frequency) is between NATO B and IEEE P
375423
This is the central processing unit (CPU) in your computer.  It is the brains of the machine and executes the programs that you run.  The MHz refers to the speed that it operates, in millions of cycles per second.  Intel Pentium is one brand, an equivalent is made by AMD (K6 and so on).  \n\nCD Rom Drive is the unit that you stick the CD into to read the program into the computer.  \n\nSounds like you are reading system requirements off a software box, in which case they are trying to tell you what your computers capabilities need to be in order to run the program.  They usually require a certain speed processor, CD Rom Drive, some hard disk space, and a given amount of system memory.  \n\nStart -> Programs -> Accessories -> System tools -> System Information should tell you the specs that your computer has.  \n\nGenerally speaking, much older computers don't run the newest software well, try to find a program that came out about the same time as the computer, or within a couple of years.  \n\nGood luck, and don't be afraid to put more details in your questions in the future.
221028
1,000 megahertz=1 gigahertz. Mega=thousand, whereas giga=million. Same rule applies for Megabyte, and Gigabyte.
810936
f = c/L, where f is the frequency in Hz, c is the speed of light in m/s, and L is the wavelength in m.\n1 angstrom = 1e-10 m, c = 300e6 m/s, so,\nf = 300e6/4000e-10 = 750e12 Hz = 750e6 MHz...that's 750 million MHz.
922623
v = f * lambda\n(where v is velocity, f is frequency and lambda is wavelength)\n\n:. lambda = v / f\nv = 3 * 10^8 m/s (speed of an electromagnetic wave)\nf = 88 MHz = 88000000 Hz\n\n:. lambda = 3*10^8 / 88000000\n                = 3.41 m\n\nie. the length of a typical radio wave is 3.41 m
510406
315mHz \n\nLove,\nKenneth Horowitz
354203
33Mhz is the speed that my memory in my laptop works at.
496724
The equivalent of the sempron 3000+ 1.8 GHz is about the Celeron D 335 2.8GHz according to this test report.\n\nhttp://techreport.com/reviews/2004q3/sempron/index.x?pg=1
451863
No, GHz is bigger than MHz for 1000 times.
683354
b. 93,300,000 Hz\n\nFM stations measure their frequencies in Megahertz (MHz)
711901
333MHZ refers to the max speed at which ur RAM can function. PC2700 is the standard to which the DDR memory is compliant with. so by putting both together on the RAM stick, the manufacturer  makes sure that u have all the info u need.
10500
I think the previous answer is geared towards channel allocation and not bit rate vs bandwidth.  I will try to clarify the issue.\n\nFirst, you must understand how FM works and how a digital signal converts to FSK (frequency shift keying).  Therefore I will use a simple example.\n\nEach bit takes up some frequency.  For example, the bit '1' could take up 5 Hz carrier frequency on the bandwidth spectrum and '0' could take up 6 Hz on the bandwidth spectrum.  Therfore, there is a seperation of 1 Hz between bits.  \n\nIf I send a '1' followed by a '0' then on a meter I should see 5 Hz freqency and then a 6 Hz frequency back to back.  We could speed this up to infinity and therefore have these 2 frequencies transmitting all data at infinite speeds.\n\nIn the real world however, a '1' may fall around the 5 Hz frequency but not exactly on it.  It may be about 5 Hz.  Why around 5 Hz and not exactly on it everytime?  Because in the real world we have noise in transmission lines.  Noise can distort the frequency causing it to be a little less than 5 Hz or a little more than 5 Hz.  Imagine if the '1' bit is transmitted but on the receiving end it ends up receieved as 5.5 Hz.  What happens now?  It's not quite 5 Hz and it's not quite 6 Hz.  Noise caused it to move 0.5 Hz away from 5 Hz.  It is in the middle of 5 Hz and 6 Hz.\n\nNow the receiving end has to decide if this is '1' or '0'.  What are the odds the receiving end gets it right?  By slowing down the speed the 1s and 0s that are transmitted , you can hold steady the 5Hz or 6 Hz for that bit.  The faster you go, the less accurate the bit will fall exactly where it should on the frequency.\n\nSo, what are some possible solutions?  Well, you can slow the frequency down or make the frequency seperation larger.  For example, make a '1' be 5 Hz and make '0' be 10 Hz.  Now you have 5 Hz seperation between bits instead of 1 Hz seperation.  Now if the frequency of 5.5 Hz is received, the receiving end can determine that this should have been 5 Hz.  It's highly unlikely it was originally transmitted at 10 Hz.  Therefore, the receiving end detects a '1'.  \n\nBut what have you lost by doing this increase in frequency seperation?  You have taken up more bandwidth.  Every cable has only so much bandwidth and if you have several channels on a cable all transmitting at the same time, you need to limit the amount of bandwidth each channel takes up or you will run out.\n\nSo, therefore, since most companies don't want to backdown on speed to the consumer (you), then they upgrade their technology (i.e. cables) to higher bandwidth cables such as fiber optic cables.\n\nI hope this helped.  Others, please feel free to expand on this topic.  I've only discussed FSK modulation.  Other techniques such as PSK, ASK, QPSK, bit stuffing, NRZI, CRC, and FEC codes can improve bit rate over channels.\n\nYou can read more about bandwidth limits.  Shannon and Hartley came up with a theorem which still holds today and is used quite extensively.  I even know someone who worked with Shannon at Bell Labs back in the day.  Shannon's theorem is C = BW*log2(1+S/N) where C is the channel capacity in bits per second,  BW is bandwidth, S is the signal power and N is the noise power.\n\nWF\nTelecom Engineer
234122
MB is megabytes, a unit of storage space.  MH is nothing unless you mean MHz which is MegaHertz which is a unit of frequency.  In other words, apples and oranges
126863
The RF waves emanating from an antenna are generated by the movement of electrical charges in the antenna. Electromagnetic waves can be characterized by a wavelength and a frequency. The wavelength is the distance covered by one complete cycle of the electromagnetic wave, while the frequency is the number of electromagnetic waves passing a given point in one second. The frequency of an RF signal is usually expressed in terms of a unit called the "hertz" (abbreviated "Hz"). One Hz equals one cycle per second. One megahertz ("MHz") equals one million cycles per second.
225117
I don't know u r so complicated this thing.\nRAM Defines By Its Capacity.Not Speed At All\nIt is a Temporary Memory.\nThe System To Be Speed We Need Optimum RAM Capacity.\n\nIn the Case Of CPU Is A Engine That Should Be Speed,\nThe Speedy CPU Will work fine with high capacity RAM\nHave u got it
1034914
The name 3400+ means that it even though it runs at a different frequency, it is equal in performance to the Intel 3.4GHz.  So a 3000+ would be equal to a 3.0 GHz, a 4200+ would be like the 4.2GHz and so on...