electrical conductivity of metals

The response of a metal to electromagnetic radiation is determined by the frequency dependent conductivity. This in turn depends on the available mechanisms for energy absorption by the conduction electrons at the given frequency. The electronic excitation spectrum in the superconducting state is characterised by an energy gap 2Eg. So we expect the AC conductivity to differ substantially from its normal state form at frequencies small compared with 2Eg Ifi, and to be essentially the same in the superconducting and normal states at frequencies large compared with 2Eg / h. The value of 2Eg /h, is typically in the range between microwave and infrared frequencies. In the superconducting state, an AC behaviour is observed which is indistinguishable from that in the normal state at optical frequencies. Deviations from normal state behaviour first appear in the infrared. At microwave frequencies AC behaviour fully displaying the lack of electronic absorption characteristic of an energy gap becomes completely developed.

define desert

Some common phenomena and applications related to refraction are as follows:Lens: Refraction is at the core of the functioning of a lens. A lens uses refraction for image formation which is used for several purposes such as magnification. Our eye is also a lens where what we see is projected on our retina. Here also refraction is used to form the image on the retina. Uses of a lens in our daily life are galore like magnifying glasses, eye glasses, contact lenses.Prism: A prism uses refraction to form a spectrum of colors from an incident beam of light.A prism can be used to break light up into its constituent spectral colours (the colors of the rainbow). Prisms can also be used to reflect light, or to split light into components with different polarizations.Mirage: A mirage is a naturally occurring optical phenomenon in which light rays are bent to produce a displaced image of distant objects or the sky. Some common examples are water bodies in a desert and wet roads on a hot sunny day.

decimal to fraction calculator

Example 5${\displaystyle \int7x^{4/3}dx=7\int x^{4/3}dx=7\frac{x^{4/3+1}}{\frac{4}{3}+1}=3x^{7/3}+\mathrm{c}}$, where $c$ is an arbitrary constant of integration. Example 6${\displaystyle \int\{3x^{2}+1+\frac{1}{x^{2}}\}dx=\int3x^{2}dx+\int1dx+\int\frac{1}{x^{2}}dx}$   $=3\cdot\frac{x^{3}}{3}+c_{1}+x+c_{2}+\frac{x^{-2+1}}{-2+1}+c_{3}=x^{3}\prime+x-\frac{1}{x}+c,$ where $c(=c_{1}+c_{2}+c_{3})$ is an arbitrary constant of integration. Note. There is no need to introduce an arbitrary constant after calculating each integral (as is done in the above example). By combining all arbitrary constants, we get a single arbitrary constant denoted by $c$ which is added to the final answer. Example 7${\displaystyle \int(2\cos2x+\frac{3}{x}-9e^{x})dx=2\int\cos2xdx+3\int\frac{1}{x}dx-9\int e^{x}dx}$ $=2{\displaystyle \frac{\sin2x}{2}+3\log|x|-9e^{x}+c=\sin2x+3\log|x|-9e^{x}+c}$, where $c$ is an arbitrary constant of integration. Example 8${\displaystyle \int\frac{(2x-3)^{3}}{x^{2}}dx=\int\frac{8x^{3}-36x^{2}+54x-27}{x^{2}}dx}$ $=8{\displaystyle \int xdx,-36\int dx+54\int\frac{1}{x}dx-27\int\frac{1}{x^{2}}dx}$  $=4x^{2}-36x+54{\displaystyle \log|x|+\frac{27}{x}+c}$, where $\mathrm{c}$ is an arbitrary constant of integration. Example 9$\int \left (3\sec 3x\tan 3x-8e^{8x}+20\sec ^2 20x+\frac {5}{\sqrt {1-x^2}} \right )dx$  $=3\cdot\frac{\sec3x}{3}-8\cdot\frac{e^{8x}}{8}+20\cdot\frac{\tan20x}{20}+5\sin^{-1}x+c$  $=\sec3x-e^{8x}+\tan20x+5\sin^{-1}x+c,$ where $c$ is an arbitrary constant of integration. Example 10 Find $y$ when $dy/dx=x^{\dot{2}}$. $dy=x^{2}dx$ and hence${\displaystyle \int dy=\int x^{2}dx}$, or,   $y={\displaystyle \frac]{3}x^{3}+c}$.

natural log rules

The base a can be changed to e.Therefore,         ex = m,Then x is the logarithm of m to the base e and it is written as,                           logem  or ln m, this is called the natural logarithms.                           Where e = 2.71828182846 (base of natural logarithm).that is ,      if logem = x    then ex = m.

formula for fahrenheit to celsius

Answer key 1: The area of the square is 289 sq cm.Answer key 2: The temperature 35 degrees Celsius to degrees Fahrenheit is 95 degrees Fahrenheit.Answer key 3: The new two angles are 40 degree.

finding the inverse of a function

Find the value of parameter $\alpha$ for which the funciion $f(x)=1+\alpha x,\alpha\neq0$ is the inverse of itself. SOLUTION Clearly, $f(x)$ is a bijection from R to itself. Now,  $fof^{-1}(x)=x$ $\Rightarrow f(f^{-1}(x))=x$  $\Rightarrow1+\alpha f^{-1}(x)=x$ $\Rightarrow f^{-1}(x)=\frac{x-1}{\alpha}$ It is given that  $f(x)=f^{-1}(x)$ for all$x\in R$   $\Rightarrow1+{\displaystyle \alpha x=\frac{x-1}{\alpha}}for\: all\: x\in R$   $\Rightarrow{\displaystyle \alpha x+1=(\frac{1}{\alpha})x+\left(\frac{-1}{\alpha}\right)}for\: all\: x\in R$   $\Rightarrow{\displaystyle \alpha=\frac{1}{\alpha}}and1=-{\displaystyle \frac{1}{\alpha}}$   $\Rightarrow\alpha^{2}=1\mathrm{and}\alpha=-1$   $\Rightarrow\alpha=-1.$

1984 chapter summaries

Force is equal to the rate of change of momentum. An unbalanced force F causes a mass m to accelerate with acceleration a given by F = ma. The impulse p of a force F acting for a time t is Ft and is equal to the change in momentum. The change in momentum is calculated from the difference between the final momentum and the initial momentum. Momentum is always conserved in collisions but kinetic energy is not necessarily conserved.The following worked examples will illustrate some of these points.ExampleA car of mass 1200 kg is brought to rest from a speed of 20 m s"1 by a constant braking force 3000 N. Calculate the retardation and the time the car takes to come to rest.F= -3000 N m= 1200 kg a = ?ExampleAn ice hockey puck of mass 0.1 kg travelling at 20 ms"1 is struck by a stick so as to return it along its original path at 10 ms"1. Calculate the impulse of the force applied by the hockey stick.Let p be the impulse produced by the Force F applied by the stick for the time t.m = 0.1 kg u = +20 m s 1 i/=-10ms_1 F=1 t = ?P = Ft = mv — mu= 0.1 x(-10) -0.1 x (+20) = - 1 -2 = -3NsThe impulse applied by the stick is -3 N s. The minus sign indicates that the direction of the impulse is in the opposite direction to the original direction of motion of the puck.ExampleA rifle pellet of mass 5 g is fired horizontally into a block of wood fixed to a model truck which runs freely on a straight track. The mass of the truck and the block is 495 g and these move with a velocity of 0.8 m s"1 when the pellet embeds itself into the wood. Calculate the initial velocity of the pellet.m = 0.005 kg u = l M = 0.495 kg v= 0.8 ms"1mu = (m + M\v 0.005 x u = (0.005 + 0.495) x 0.8_ 0.5 x 0.8 u 0.005u- 80 ms"1The initial velocity of the pellet is 80 m s-1.ExampleA bullet of mass 0.01 kg travelling horizontally at 100 ms"1 penetrates a fixed block of wood and comes to rest in 0.02 s. Calculate (a) the distance of penetration x of the bullet into the wood and (b) the average retarding force exerted by the wood on the bullet.(a)u= 100ms"1 f=0ms_1 a = ? /=0.02s v=u +at 0 = 100 + a x 0.02 0.02a = -100 a = -100/0.02 a = -5000 ms"2v2 = u2 + 2 ax 0= 1002-2x5000* 10 000 x = 10 000 *=lmThe bullet penetrates 1 m into the block (note the deep penetration).(b)F=? m = 0.01 kg a = -5000 ms"2 F= maF= 0.01 x (-5000) F= -50 NThe average retarding force exerted by the wood is -50 N.