Please explain in details

Q-1: Consider the following fraction of memory and the CPU registers and their contents. Memory Address 50001 5004 5008 5008 5012 50125016 5020 5024 5028 5032 5036 2A66005612568790 5040 5044 5687124500005678 2A660OS6 00000057 00000057 55446699 5048 5052 5056 5060 5064 2660056 12548763 0055968000000002A660056 Memory Content 00225500 2A660056225SOOAA 5068 5072 5076 EF 253147ABABABAB 25896314 56789-402 82355FF in 16 bit - 0000 5008 Register Register Content Instructions: Name $SO BF340F55 LW $t0, 12(551) $to = OOO000 57 4 $S1 5008 ADD $t2, $to, $t1 $ +2 = A1232 58B + obce0057 00000000 66775544 sw $t2, 24($51) $to 12328585 $t1 A123258B $to put it in address 24 (851) $t2 00110011 Assume that you have an integer array of 10 elements whose base address is stored in register $51. Run the above instructions and fill out the following table. Big Endian Little Endian Memory Content Memory MSB LSB MSB LSB Register $51 oo 00 5008 Register $51 08 50 00 00 Memory location 5020 oo 00 00 57 Memory location 5020 57 00 00 00 Register Sto 00 00 00 5+ Register Sto 57 00 00 00 Register $t2 IAI 23 25 E2 Register $t2 E2 25 23 Ai Memory Location 5032 AL 23 25 2 Memory Location 5032 E2 | 25 | 23 A