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Are you able to take apart a Playmobil man?

no


How do you take a plug apart?

Not possible without breaking it.


If you drop a rock from a great height about how fast will it be falling after 5 seconds neglecting air resistance?

The object will be falling at 49 m/s.This is solved by multiplying the force of gravity (9.8 m/s) by the time you're calculating (5s).


How do you take apart the dash to put a aftermarket in?

youtube com/watch?v=dQw4w9WgXcQ


What are the instruction of ias computer?

InstructionTypeOpcodeSymbolicRepresentationDescriptionData transfer00001010000010010010000100000001000000100000001100000100LOAD MQLOAD MQ,M(X)STOR M(X)LOAD M(X)LOAD -M(X)LOAD |M(X)|LOAD -|M(X)|Transfer contents of register MQ to the accumulator ACTransfer contents of memory location X to MQTransfer contents of accumulator to memory location XTransfer M(X) to the accumulatorTransfer -M(X) to the accumulatorTransfer absolute value of M(X) to the accumulatorTransfer -|M(X)| to the accumulatorUnconditionalbranch0000110100001110JUMP M(X,0:19)JUMP M(X,20:39)Take next instruction from left half of M(X)Take next instruction from right half of M(X)Conditionalbranch0000111100010000JUMP+M(X,0:19)JUMP+M(X,20:39)If number in the accumulator is nonnegative, take next instruction from left half of M(X)If number in the accumulator is nonnegative , take next instruction from right half of M(X)Arithmetic0000010100000111000001100000100000001011000011000001010000010101ADD M(X)ADD |M(X)|SUB M(X)SUB |M(X)|MUL M(X)DIV M(X)LSHRSHAdd M(X) to AC; put the result in ACAdd |M(X)| to AC; put the result in ACSubtract M(X) from AC; put the result in ACSubtract |M(X)} from AC; put the remainder in ACMultiply M(X) by M(Q); put most significant bits of result in AC, put less significant bits in M(Q)Divide AC by M(X); put the quotient in MQ and the remainder in ACMultiply accumulator by 2 (i.e., shift left one bit position)Divide accumulator by 2 (i.e., shift right one bit position)Address modify0001001000010011STOR M(X,8:19)STOR M(X,28:39)Replace left address field at M(X) by 12 right-most bits of ACReplace right address field at M(X) by 12 right-most bits of AC