in the 8051, how many bytes of rom space are assigned to the reset interrupt, and why?
375500
The reset interrupt is at location 0000H and is 3 bytes long. All other interrupts are 8 bytes long. Its at location 0000H because that's where the processor starts executing code on power up. Why 3 bytes? The architecture was designed to have all of the interrupts located at the beginning of ROM. For a reset interrupt, the code in the interrupt will be a LJMP (long jump) to another location in ROM, which takes 3 bytes.
275553
(m)\n\nThis is system memory that is simulated by the hard drive. When all the RAM is being used (for example if there are many programs open at the same time) the computer will swap data to the hard drive and back to give the impression that there is slightly more memory.
733285
www.crucial.com\nsome of the memory may be diverted over for video memory.  There may be a setting in the BIOS.
217756
I am sorry...7490 is a decade counter, NOT a 7-segment driver...\n\nIn the 7490, pins 2&3 are internally NAND'd so that a reset occurs in BOTH internal counters only when  BOTH pins are brought to a HIGH state...I'd advise against leaving EITHER reset unterminated.....Tying them together will reset both counters at once.\n\nExternally, there is NOTHING wrong with ORing reset signals from several sources...Just consider the 'fan out' and 'race' conditions that might apply according to your design requirements.\n\nDown load a pdf datasheet for 7490 at the resource below.
974636
find the motherboard online and check how much memory it can hold
520485
Run the Crucial Memory Advisor - http://www.crucial.com/promo/index.asp?prog=desktop2
438803
I've actually pondered this myself when taking digital circuit courses... The solution , i THINK, involves having not just one large memory unit , but instead, breaking the memory into 4 or 8 or 16 or more parts and using some kind of chip-select mechanism (using a multiplexer)
182732
SM581000A-7 is listing as a SIMM memory module. It was used in PC's back in the 80286 and 80386 days.
1098340
According to the HP Maintenance and Service Manual, there are 2 slots for PC2100 DDR-266 SDRAM modules which support up to 1GB (1024 MB) SDRAM using 512 MB modules
628097
Imagine a block of memory from 0 to N. A cyclic buffer fills from 0 towards N, and when it reaches N it starts back at 0 again.\nThis kind of buffer needs to be "serviced" (have the data removed) before new data overwrites data that hasn't been removed yet.\nThere is normally an "in" pointer for incoming data, and an "out" pointer for outgoing data, If the "out" pointer reaches the "in" pointer, the buffer is empty again. If the "in" pointer reaches the "out" pointer, you've overflowed the cyclic buffer!
570968
Reseat the memory in the slot, or try switching them around. Sounds like a bad connection.
1153480
You could try dividing by zero, or writing in a nonexistant memory location, but the Operating system won't let you do those things.  \n\nYour best bet might be to pull the power plug.\n\nNote that you may lose data!
660413
115,200 bits per second
86332
You should visit to www.intel.com\nThere are so many documents talking about memory management.
1144484
It's tellin you that you have a 256 MB module in each U5 and U6 slots. making a Total of 512 MB.
208512
double type is probably memory (ram) too low because not enough,at least 512 not shared by graphics on board MB is nominal,floppy issue is not important because operating system does not utilize to run programs,is save only or older programs upload.if do not need floppy initially ,boot,do the f key row to access bios,usually 4 7 9 11 depends on system,push all at boot.at bios go to BOOT SEQUENCE and select cd or dvd drives as first boot,and floppy as turned off.reboot and decide if okay,run program such as www.iolo .coms system mechanic trial to clean up system,free memory and give analysis,or tell me i am of rack and missed question,good luck.......
194032
world's first microprocessor, the 4004, where the '4's were used to indicate that the device had a 4-bit data path. The 4004 was part of a four-chip system which also consisted of a 256-byte ROM, a 32-bit RAM, and a 10-bit shift register. The 4004 itself contained approximately 2,300 transistors and could execute 60,000 operations per second. The advantage (as far as Hoff was concerned) was that by simply changing the external program, the same device could be used for a multitude of future project