Thursday, December 22, 2016

Fundamental Physics Quick Revision 11th Standard 101-1a

101-1a


Here are some fundamental mathematical basics of 11th Standard Physics for your quick reference

This one 101-1a will cover all the basic formulae you will need to solve your Kinematics problems and some for basic unit conversions.

SI Prefixes*

Factor  Prefix  Symbol
1024    yotta        Y
10
21    zetta         Z
10
18    exa           E
10
15    peta          P
10
12    tera          T
10
9      giga         G
10
6      mega       M
10
3      kilo          k
10
2      hecto        h
10
1      deka        da



Factor   Prefix    Symbol
10–1       deci        d
10
–2       centi       c
10
–3       milli       m
10
–6       micro     mu 
10
–9       nano       n
10
–12     pico        p
10
–15     femto      f
10
–18     atto         a
10
–21     zepto      z
10
–24     yocto      y


*In all cases, the first syllable is accented, as in ná-no-mé-ter.

 

 

Trigonometric Identities:


Reciprocal identities

displaymath161
Pythagorean Identities
displaymath162
Quotient Identities
displaymath163
Co-Function Identities
displaymath164
Even-Odd Identities
displaymath165
Sum-Difference Formulas
displaymath166
Double Angle Formulas
align99
Power-Reducing/Half Angle Formulas
displaymath167
Sum-to-Product Formulas
displaymath168
Product-to-Sum Formulas
displaymath169

Basic Derivatives and Integrals:

 

 

Newton's generalized binomial theorem

In this generalization, the finite sum is replaced by an infinite series. In order to do this, one needs to give meaning to binomial coefficients with an arbitrary upper index, which cannot be done using the usual formula with factorials. However, for an arbitrary number r, one can define
{r \choose k}={\frac {r\,(r-1)\cdots (r-k+1)}{k!}}={\frac {(r)_{k}}{k!}},
where (\cdot )_{k} is the Pochhammer symbol, here standing for a falling factorial. This agrees with the usual definitions when r is a nonnegative integer. Then, if x and y are real numbers with |x| > |y|, and r is any complex number, one has
{\displaystyle {\begin{aligned}(x+y)^{r}&=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}\\&=x^{r}+rx^{r-1}y+{\frac {r(r-1)}{2!}}x^{r-2}y^{2}+{\frac {r(r-1)(r-2)}{3!}}x^{r-3}y^{3}+\cdots .\end{aligned}}}
When r is a non negative integer, the binomial coefficients for k > r are zero, so this equation reduces to the usual binomial theorem, and there are at most r + 1 nonzero terms. For other values of r, the series typically has infinitely many nonzero terms.
For example, with r = 1/2 gives the following series for the square root:
{\sqrt {1+x}}=\textstyle 1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+{\frac {7}{256}}x^{5}-\cdots
Taking r=-1, the generalized binomial series gives the geometric series formula, valid for |x|<1:
(1+x)^{-1}={\frac {1}{1+x}}=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots
More generally, with r = −s:
{\frac {1}{(1-x)^{s}}}=\sum _{k=0}^{\infty }{s+k-1 \choose k}x^{k}\equiv \sum _{k=0}^{\infty }{s+k-1 \choose s-1}x^{k}.
So, for instance, when s=1/2,
{\frac {1}{\sqrt {1+x}}}=\textstyle 1-{\frac {1}{2}}x+{\frac {3}{8}}x^{2}-{\frac {5}{16}}x^{3}+{\frac {35}{128}}x^{4}-{\frac {63}{256}}x^{5}+\cdots

 

Vector Product of Vectors

The vector product and the scalar product are the two ways of multiplying vectors which see the most application in physics and astronomy. The magnitude of the vector product of two vectors can be constructed by taking the product of the magnitudes of the vectors times the sine of the angle (<180 degrees) between them. The magnitude of the vector product can be expressed in the form:  and the direction is given by the right-hand rule. If the vectors are expressed in terms of unit vectors i, j, and k in the x, y, and z directions, then the vector product can be expressed in the rather cumbersome form:
 

Vector Product, Determinant Form

The vector product is compactly stated in the form of a determinant which for the 3x3 case has a convenient evaluation procedure:
Once the scheme for determinant evaluation is familiar, this is a convenient way to reconstruct the expanded form:






Author & Editor

Editor,Tricorp Web Designs

0 comments:

Post a Comment