wolfgang+gassmann

question 1 We assume before the car hits the wall all of its energy is kinetic energy and after the collision it is all spring energy Thus K.E before the crash equals spring energy after the crash .5mv^2=.5kx2 mv^2=kx^2 m=kx^2/v^2 question 2 we know at any point in the springs motion that K.E+ spring energy= total energy so when all of the energy is in spring energy total energy= spring energy. total energy in the system is constant so K.E+spring energy(at halfway point)=spring energy( when the spring is fully compressed/extended) K.E=spring energy(hf)-spring energy(fc) mv^2=kx 1 ^2-kx 2 ^2 m= (kx 1 ^2-kx 2 ^2)/v^2 ii T=2pi*sqrt(m/k) input values from previous problem to get answer iii am=xm+omega^2 am=xm+sqrt(m/k)^2 am=xm+m/k question 3 vm=xm+omega omega=sqrt(m/k) vm=xm+sqrt(m/k) ii mv^2+kx 1 ^2=kx 2 ^2 where x2 is at the end point and x1 is at .04 meters from the equilibrium mv^2=kx 2 ^2-kx 1 ^2 v=sqrt((kx 2 ^2-kx 1 ^2)/m) iii f=kx ma=kx a=kx/m Week 5 Imagine a triangle fomed by the string with its sides being 3L/4, .5L, and .5L Now cut the triangle in half so that you have a triangle with the hyponunues= .5L and the other leg= 3L/8 using this and trig you can solve fo the angle which is the angle at which the string hangs cos^-1(.75) knowing v=sqrt(F/u) and F=mg we get v^2=mg/u acounting for the angle and the exsistance of two strands of string we get v^2=2sin(cos^-1(.75))*m*g/u now solve for mass 2 T=2pi*sqrt(L/g) L=g(T/(2pi))^2 u=m(string)/L and v=Sqrt(m(hanging object)/u) for both porblems dont forget to convert everything to SI units
 * Week 3**

01. The “red shift” of radiation from a distant galaxy consists of the light known to have a wavelength of 434 nm when observed in the laboratory, appearing to have a wavelength of 462 nm. What is the speed of galaxy in the line of sight relative to the Earth? The speed of light is 2.99792 × 10^8 m/s. wavelength'=wavelength+velocity of source*period period=lambda/velocity wavelength'=wavelength+velocity of source*wavelength/velocity (λ'/λ-1)v of light=v of source don't forget to convert nanometers to meters
 * week 6**

02. The sound of a man shouting at the top of his lungs from a rather large distance away from your ear has loudness of only 20 decibels. What would be the decibel level of four men shouting at the top of their (equally powerful) lungs from the same distance away from you ear? Assume that there is no interference from superposed waves. 1 man has x power so 4 men have 4x power so additional sound level=10*log(4x/x) 20+additional sound level= new sound level

Professor's Note : Very good Job !! - Proud of you :)

week 8 polarization describes the direction which the energy of a wave moves, in a polarized wave they all move in the same direction, in a non polarized wave they move in all sorts of directions 1. Initially unpolarized light is sent into a system of three polarizing sheets whose polarizing directions make angles of Ø1 = Ø2 = Ø3 = 50⁰ with the directions of the y axis. What percentage of the initial intensity is transmitted by the system? (hint: be careful with the angles) We use the half rule knowing we are converting unpolarized light to polarized light so I=.5I o. This I represents polarized light that makes an angle of 50 degrees with the y axis all the other sheets polarize at this angle as well meaning the light passes through them unphased as it has already been polarized at 50 degrees meaning I=.5I o would give you the answer with some algebra and converting to percentages we could use I=cos^2(theta)Io but theta the angle between the light and the polarizing direction would be equal to 0 as 50-50=0 which would make cos(theta)=1 and I=Io Though this only works because all of the sheets polarize in the same direction,you would use the equation I=cos^2(theta)Io when the polarizing directions are different week 09

1. A fruit fly of height H sits in front of a lens (say lens 1) on the central axis. The lens forms an image of the fly at a distance d = 20 cm from the fly ; the image has the same orientation as the fly and a height = 2 H.

a) What are the focal length of the lens and the object distance of the fly?

20cm=abs(p)+abs(i) the image is up right so m is positive m=2 2=-i/p with these to equations you can solve for p of the fly and i as well. then using 1/f=1/p+1/f you can solve for f and solve for the focal length

Fly then leaves lens 1 and sits in front of a second lens (say lens 2), which also forms an image at d = 20 cm from the fly and has the same orientationas of the fly, but now the image height is 0.5 H

b) What are the focal length of the second lens and the new object distance? 20cm=abs(p)+abs(i) the image is up right so again m is positive m=.5H'/H m=.5 .5=-i/p use the 2 above equations to solve for p and i then using 1/p+1/i=1/f you can find the focal length of the lens

2. When the rectangular metal tank in the figure, is filled to the top with an unknown liquid, Observer O with eyes level with the top of the tank, can just see corner E. Aray that refracts toward O at the top surface of the liquid is shown. If D = 85.0 cm and L = 1.1 m, what is the index of refraction of the liquid? L=1.1 D=.85 using D and L as legs of a right triangle it is posible using trig to solve for the angle with respect to the vertical knowing that the second angle is 90 because the light refracts to the horizontal we can say angle1=tan^-1(L/D) and knowing n2 is of air so=1 n1*sin(tan^-1(L/D))=1*sin(90) n1=1/sin(tan^-1(L/D)

Professor's Note on Week 10 : No work was found !!

a ) In a double slit arrangement, the slits S1 and S2 are illuminated with light whose wavelength is 643 nm. The slit separation is 0.15 mm, slit to screen distance is 145 cm. What is the spacing between the bright fringes on the viewing screen?

b) A thin sheet of transparent plastic of thickness t = 0.48 μm with a refractive index of n = 1.5 is placed over only the upper slit.

As a result, the central maximum of the interference pattern moves upward a distance y, find y.

assuming sin=tan=y/D as this problem describes a right triangle we can use the double slit equation dsin=m*lamda by subbing in tan for sin we get d*y/D=m*lamda solving for y we get y=m*lamda*D/d making delta y=2 *lamda*D/d -1*lamda*D/d

b N is the total number of wavelengths along the path of light from the source to the screen, r is the length of the path which the light travels N1=N t +Nair=(r1-t)/lamda+t/lamda of light going through plastic N2=r2/lamda central max is where N1=N2 so r2/lamda=(r1-t)/lamda+t/lamda of t knowing r2=r1 +dsin (r1+dsin)/lamda=(r1-t)/lamda+t/lamda of t (r1+dsin)/lamda=(r1-t)/lamda+t*n/lamda (dsin+r1)/lamda=(t(n-1)+r1)/lamda simplyfy dsin=t(n-1) tan=y/L y/L=t(n-1)/d solve for y

A rectangular plate has an area of 95 cm². If the temperature increases by 112 ºC, calculate the increase in the area. Use 5 × 10 − 6 (ºC)-¹ as an average coefficient of linear expansion, and ignore terms of second order.

using the equation deltaL=L*deltaT*alpha and the concept that to convert the liner coefficient of expansion to the area coefficienct of expansion you multiply by 2 we get delta V=V*deltaT*2*alpha you could also assume each side of the rectangle is sqrt(95) as if it is a square find the linear change then calculate the area of the expanded square using the new values and take the difference between the new and the old, you get the same answer

2. A quantity of ideal gas at 10 ºC and 100 kPa occupies a volume of 2.50 m³. (a). How many moles of the gas are present? If the pressure is now raised to 300 kPa and the temperature is raised to 30 ºC, how much volume does the gas occupy? Assume there are no leaks. PV=nRT for part a simply solve for n being sure to convert celcius to kelvin, kpas to pascals to keep with the SI units so PV/(RT)=n for part b you can assume the number of moles stays constant so PV/(RT)=n=P 2 V 2 /(RT 2 ) R cancels out PV/(T)=n=P 2 V 2 /(T 2 ) so simply solve for V2