BODMAS BOMDAS
Many people have been having trouble with the use of BODMAS, BOMDAS, etc.
Example: 7-3+4 = ? .....(1)
Now if addition comes before subtraction we should add the 3 and the 4 so that (1) becomes 7-7, which equals 0.
It is not much good saying that we must add the 7 and 4, because there is no indication that this is what it means. Remember that when we talk of addition we are talking of TWO operands (the 3 and the 4) and ONE operator (the + sign between the 3 and 4).
Now the correct answer is that 7-3+4 is 4+4 (the subtraction was done first - but BODMAS says do addition first).
Here I will bluntly say that BODMAS is not correct if used in this way. If you underline the DM part and the AS part and say that D and M have equal precedence and that the A and S have equal precedence you are halfway there. If you also say that YOU MUST READ FROM LEFT TO RIGHT you will get it correct by following this: In 7-3+4 read from left to right and do addition or subtraction as you come across each.
7-3+4=4+4=8 is correct.
More comprehensively:
1)read from left to right
2) exponentiation comes first
3) then multiplication and division (equal precedence)
4) then addition and subtraction (equal precedence)
5) If there are brackets start with the innermost brackets and apply the above rules within the brackets.
Further note: With 7-3+4 you have the subtraction 7-3 with the two operands (7 and 3) and the operator (-) and addition 3+4 with the two operands (3 and 4) and the operator (+).
People who study computer science will be familiar with precedence of operators.
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Physics hints:
Equations of motion
v = u + at ....(1)
v²
= u²
+ 2as .....(2)
s = ut+½at² ......(3)
Can you see that there is no equation having the variables (see note 2) a, s, t and v in? How would you solve a problem involving a, s, t, and v? Say you had to find t and you were given a, s and v. Well you could find u using equation (2) since you are given a, s and v. Then you could find t using (1) or (3). (1) is easiest.
NOTE: Each equation involves 4 variables (see note 2) and you need to know the values of 3 of them to solve a problem using any one of the equations.
NOTE 2: I speak of variables in the sense that the four quantities vary from problem to problem. In fact the equations are derived assuming CONSTANT acceleration. Within a problem u is the initial velocity and remains constant and acceleration a is constant. The distance s changes as does t as the object moves and time progresses. When doing problems it is best to refer to them as QUANTITIES and not variables.
Why does the current remain the same in any part of a conductor (say a long wire)?
Answer: If the current into a portion of the wire was more than the current out of that portion there would be a build up of charge. Say we had 2 coulombs built up in that portion of wire (remember 1 amp means one coulomb flowing past a point in one second), then we would have huge repulsion (remember Coulomb's law?) If less charge went in than went out we would also have an excess of positive charge (usually current is a flow of electrons within a wire, but one can have positive charges flowing in other situations).
Coulomb's law:NOTE: Each equation involves 4 variables (see note 2) and you need to know the values of 3 of them to solve a problem using any one of the equations.
NOTE 2: I speak of variables in the sense that the four quantities vary from problem to problem. In fact the equations are derived assuming CONSTANT acceleration. Within a problem u is the initial velocity and remains constant and acceleration a is constant. The distance s changes as does t as the object moves and time progresses. When doing problems it is best to refer to them as QUANTITIES and not variables.
Why does the current remain the same in any part of a conductor (say a long wire)?
Answer: If the current into a portion of the wire was more than the current out of that portion there would be a build up of charge. Say we had 2 coulombs built up in that portion of wire (remember 1 amp means one coulomb flowing past a point in one second), then we would have huge repulsion (remember Coulomb's law?) If less charge went in than went out we would also have an excess of positive charge (usually current is a flow of electrons within a wire, but one can have positive charges flowing in other situations).
F = (kq1q2)/r2
Above is the mathematical statement. In words: The force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The constant k is not like the constants discussed above, but is always the same from problem to problem. It is called Coulomb's constant or called the electric force constant or electrostatic constant.
If you had to make r the subject of the formula how would you go about it? You could multiply both sides by r2 and then divide both sides by F and then take the square root of both sides (or multiply both sides by r2/F and take the square root of both sides).
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