Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
The complete program is as follows:
m_str = input('Input m: ')
mass = float(m_str)
e = mass * 300000000**2
print("e = ",e)
Explanation:
This is an unchanged part of the program
m_str = input('Input m: ')
This converts m_str to float
mass = float(m_str)
This calculates the energy, e
e = mass * 300000000**2
This is an unchanged part of the program
print("e = ",e)
C. system software
if it means drivers and so on
Answers- My computer, My Documents and Recyle bin.
2,3,4 option
<em>https://todo-anime.com/</em>
