MReid

__**Week eleven (28th November, 2011) **__
 * 1. 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.**

Step 1: assuming the plate is a square, find lenght from L=sqrt(95)

Step 2: Δ L= L*α*Δ T, where ΔL= change in length, L= length, α= 5*10^-6

Step 3: add change in length and length then square it to get new area,

Step 4: subtract final area from initial area to get the change,


 * 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?**

Step 1: convert kPa to Pa and convert Celsius to Kelvin

Step 2: PV=nRT, where P=pressure=100kPa, V=volume=2.5m^3, n=number of moles=?, R=constant=8.31J/molK, T=temperature=10 degrees Celsius

Step 3: solve for n


 * 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.**

Step 1: pressure and temperature have changed, n remains the same, and the new volume is unknown. Therefore, use PV=nRT and solve for V

__Week ten (14th of November, 2011)__


 * 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? **


 * Step 1: Draw diagram**


 * Step 2: make sure that the units of measurements are the same**


 * Step 3: λ=yd/L therefore solve for y= the spacing between the bright fringes**


 * 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. **


 * Step 1: make sure that the units of measurements are the same**


 * Step 2: y=L*t(n-1)/d**


 * Professor's Note: Matt, you are giving me the final equation that you derived elsewhere. Please give me the steps and also note these equations are not general equations. **

Week eight ( 18 October, 2011)


 * 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)**


 * Step 1: By Malus' Law we know that when the unpolarized light enters the first polarizing sheet, it looses half its intensity. I1 = Io/2**


 * Step 2: Also by Malus' Law, I2= I1*cos^2(Ø2-Ø1) and I3= I2*cos^2(Ø3-Ø2)**


 * Step 3: the three equations combined can be rewritten as I3= (Io/2) * cos^2(Ø2-Ø1) * cos^(Ø3-Ø2)**


 * Step 4: solve for the ratio (I3/Io) then multiply that value of that ratio by 100 to get the percentage transmitted.**


 * Week seven ( 10th October, 2011) **


 * 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.**


 * note: 1nm = 1 * 10^-9 m**


 * Step 1: identify given and unknown: v=speed of light=2.99792 * 10^8, vs=speed of galaxy=?, λo= wavelength when observed= 434nm, λs=apparent wavelength =462nm, fo=frequency when observed= ?, fs=apparent frequency= ?**


 * Step 2: use v=fλ to solve for f o and fs, using speed of light in both cases.**


 * Step 3: use Doppler effect equation, fo=fs[1/(1+(vs/v))], and solve for vs**


 * 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.**


 * β1=decibel level for 1 man= 10log(I/Io)= 20dB, therefore β4=decibel level for 4 men= 10log(4I/Io), by log rules, = 10log(I/Io)+10log(4)= 20+10log(4)=...**


 * Professor's Note : good job on both .... way to go Matt ! **

Third week (12th September, 2011)

1.** An automobile having a mass of 1000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring of constant 5 × 10^6 N/m and compresses 3.16 cm as the car is brought to rest. What was the speed of the car before impact, assuming no energy is lost during impact with the wall? **


 *  Step 1: identify given and unknown: m=mass=1000kg, k=constant =5*10^6N/m, x=distance of compression=3.16cm, v=speed=? **


 * Step 2: make any necessary conversions (cm ---> m) **


 * Step 3: Energy is conserved, therefore using the equation KE = (1/2)mv^2 = PE = (1/2)kx^2, solve for v **

2**. A block of unknown mass is attached to a spring of spring constant 6 .5 N/m and undergoes simple harmonic motion with an amplitude of 10 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 30 cm/s.**
 * i. Calculate the mass of the block. **


 * Step 1: identify the given and unknown: k=spring constant=6.5N/m, x(max)=max amplitude=10cm, v=speed=30m/s, m=mass=? **


 * Step 2: using the equation Omega= v/x(max) = sqrt(k/m), solve for m **


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">ii. Find the period of the motion. **


 * <span style="font-family: Arial,Helvetica,sans-serif;">Step 1: using the information from the first part and the equation Omega=2pi/T, solve for T, the period of the motion. **


 * <span style="font-family: Arial,Helvetica,sans-serif;">iii.Calculate the maximum acceleration of the block. **


 * <span style="font-family: Arial,Helvetica,sans-serif;">Step 1: Using the information from the first parts and the equation, a(max)=Omega^2*x(max), find a(max), the maximum acceleration. **

3. **A 0.4 kg block attached to a spring of force constant 12 N/m oscillates with an amplitude of 8 cm.**
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">i. Find the maximum speed of the block. **


 * Step 1: m=mass=0.4kg, k=spring force constant=12N/m, x(max)=max amplitude=8cm**


 * Step 2: make any necessary conversions (cm ---> m)**


 * Step 3: using v(max)=x*Omega, and Omega=sqrt(k/m), solve for v(max)**


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">ii. Find the speed of the block when it is 4 cm from the equilibrium position. **


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Step 1: make conversion (cm ---> m) **


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Step 2: using the equation x(t)=x(max)*cos(Omega*t), solve or t **


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Step 3: using the equation v(t)=x(max)*Omega*sin(Omega*t), solve for v **


 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 110%;">iii. Find its acceleration at 4 cm from the equilibrium position. **


 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Step 1: using the equation a=-Omega^2*x(t), solve for acceleration a **
 * Professor's Note: Good job, overall **

__Second week : (05th of September, 2011)__ 1. **A log-raft was made by lashing six logs together, where each log has a diameter 0.25 m and length 1.90 m. How many children can safely afloat on the raft in sea water, if the average weight of a child is 200 N? Density of the logs is 800 kg/m3 and sea water is 1024 kg/m3.**

Step 1: find the total volume of all the logs which equals: (pi)(r^2)(length of logs)(number of logs)

Step 2: find the total mass of the logs which equals: (Volume of logs)(density of logs)

Step 3: find the maximum mass (raft and passengers) which equals: (Volume of logs)(density of sea water)

Step 4: find the maximum passenger weight (in Newtons) which equals: [(maximum mass) - (mass of logs)] * (acceleration due to gravity)

Step 5: estimate the amount of children that can board safely which equals (or is less than): (maximum passenger weight) / (average weight of a child)

2. **A cube of wood whose edge is 0.015 m is floating in a liquid (in a glass beaker) with one of its faces parallel to the liquid surface. The density of wood is 550 kg /m³, that of liquid is 1240 kg/m³. How far below the liquid surface is the bottom face of the cube?**

Step 1: set up the equation: (Volume of cube submerged in liquid)(density of liquid)(acceleration due to gravity) = (Volume of cube)(density of cube)(acceleration due to gravity)

Step 2: expand Volumes: (Volume of cube submerged in liquid) = (cube edge^2) * h. Where "h" is the length of the cube that is submerged. (Volume of cube) = (cube edge^3)

Step 3: using the equation, solve for h**:** (cube edge^2) * h * (density of liquid)(acceleration due to gravity) = (cube edge^3)(density of cube)(acceleration due to gravity)

OR

The ratio of the density of the cube to the density of the liquid is equal to the ratio of the submerged length of the edge to the full length of the edge. Therefore solve for the submerged length, h: (density of cube) / (density of liquid) = h / (full edge length) Professor's Note : Excelent work, wonderful way to start the semester.. keep up the good work** <span style="display: block; font-family: 'times new roman',times,serif; font-size: 90%; height: 1px; left: -40px; overflow: hidden; position: absolute; top: -25px; width: 1px;">A block of unknown mass is attached to a spring of spring constant 6 .5 N/m and undergoes simple harmonic motion with an amplitude of 10 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 30 cm/s.