Cole-Murphy


 * __Final week__**

**Fourteenth and fifteenth weeks (Nov. 29th 2010 and Dec 6th 2010)**

**1. Three moles of a certain diatomic molecular gas are heated at constant pressure from 300 K to 400 K. The heat capacity of this gas under constant pressure is 28.8 J/mol · K and the universal gas constant is 8.31451 J/mol · K. Calculate the heat transferred to the gas. Answer in units of J.**

**__ GIVEN: __**

**Cv=(3/2)R** **Q=n*Cv*(Tf/Ti).**

**STEPS:** **1. Use the first equation to solve for Cv.** **2. Use the Cv you solved for in the second equation and plug in Ti and Tf to solve for Q**

**2. A vessel with a capacity of 5 L contains 0.125 mol of an ideal gas at 1.5 atm. The value of Boltzmann’s constant is 1.38066 × 10−23 J/K, and the universal gas constant is 8.31 J/K · mol. What is the average translational kinetic energy of a single molecule? Answer in units of J.**


 * GIVEN: **
 * PV=nRT **
 * T=(PV)/(nR) **
 * K avgerage=(3/2)k*T **

**STEPS:** **1.Take the first equation amd use algebra to change it into the second equation.** **2.Plug in the given values to solve for T.** **3.Then use the third equation to solve for K average.**

**3. A 50.0 g block of copper whose temperature is at 150 F is placed in an insulating box with a 100 g block of lead whose temperature is 100 F (a) What is the equilibrium temperature of the two-block system?**

**GIVEN:**

**S=m*c*ln(Tf/Ti)**

**STEPS** **1.Convert given to all proper units.** **2.Use algebra to get the equation mc*Cc*(Tf-Tic)+ml*Cl*(Tf-Til)=0** **3.Solve for the temp when the Cc=386 and Cl=128** **4.Plug these values into the given entropy derivative.** **5.Solve for the S of the copper and lead block and add them together.**


 * Definitions: **
 * Q = absorbed or released heat **
 * m1 =mass of copper and m2 = mass of lead **
 * c1 =specific heat of copper and **
 * c2 = specific heat of lead **
 * DT=change in temperature = (Tf - Ti) **
 * At the equilibrium, Q (absorbed) + Q (released) = 0 **

( **b) What is the change in the internal energy of the system?(c) What is the change in the entropy of the system? 4. A 600 W carnot engine operates between constant temperature reservoirs at 100 C and 60 C.(a) What is the rate at which energy is taken in by engine as heat(b) What is the rate at which energy is exhausted by the energy as heat?**


 * GIVEN: **
 * Ec=1-(Tc/Th) **
 * Ec=W/Qh **
 * P=W/t **
 * Ec=(Qh-Qc)/Qh. **


 * STEPS: **
 * 1.Use algebra with the given equations to derive Ec=(W/t)*(Qh/t) **
 * 2.Because of equation 3, Ec=P/(Qh/t) , and solve for Qh/t **
 * 3.Plug the Ec and Qh values into the last given equation to solve for Qc **

**__ 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. **

**GIVEN:** Delta L=L* Alpha* Delta T

**STEPS:** 1.Take the square root of 95 cm^2. 2 Use the above equation and use the square root of 95 at the original length 112 as the change in T and 5*10^-6 as alpha. 3. Apply this chang in L to the original l and square it. 4.subtract this number from the original area to get the change in area


 * 2. A quantity of ideal gas at 10 ºC and 100 kPa occupies a volume of 2.50 m³ . **

**GIVEN:** **PV=nRT**


 * (a). How many moles of the gas are present? **

**STEPS:** 1. Plug in the given values for Pressure Temp in kelvin, and volume as well as the gas constant. 2. Solve for n.


 * (b). 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. **

**STEPS:** 1. Manipulate the equation into V=nRT/P 2. Plug in the new values for T and P and the gas constant and the n solved for in part a. 3. 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? ====

==== 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. ====

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

STEPS: 1. All of the needed imformation is given in the problem just make sure that all distances are converted to nanometers. 2. The y value given in the equation is the difference in y

__** Week nine ( 24th October, 2011) **__
 * 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? **

GIVEN: M=2 Abs(p)+Abs(i)=20 f=? i=? p=?

STEPS: 1. We can say that Abs(p)+Abs(i)=20 and that -i/p=2, we can solve both of these as a system and then get both i and p 2. WE also know that 1/f= 1/i plus 1/p, with this we can solve for f and we already have p.


 * 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? **

**3. The same steps can be taken as before but this time with the new numbers,**


 * 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? **

**GIVEN:** **n*sin (theta)=n*sin (theta)**

**STEPS:** **1. Use the given values for the dimensions of the box to salve for the angles.** **2. Once the angles are solved for use the above equation to find the index of refraction.**


 * Week eight ( 18 October, 2011)**


 * Define polarization**

**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)**

**STEPS:** 1.Use figure 33-40 in the book to get an idea of what the situation looks like. 2.From the Book we can gather that I=Io*cos^2(50) 3.This will repeat it self and form a system:

I1=Io*cos^2(50)

I2=I1*cos^2(50)

I3=I2*cos^2(50)

4.This system can be rearranged to be I3=(cos^2(50))3 5. With this the percentage can be solved for.


 * Week 7**


 * 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. **

**GIVEN:** V=f *Lambda f(observed)=f(source)[1/(1+(Vsource/Vsound)]

**STEPS:** 1. We know the speed of light as well as the wavelength of the waves, and with the given equation V=f* lambda we can solve for their frequencies. 2. Now that we have both the oberved frequency we can use the equation f(observed)=f(source)[1/(1+(Vsource/Vsound)] and solve for the velocity of the source.


 * 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. **

**GIVEN:** Beta = 10log(I/I_o)

**STEPS:** 1. I is now going to be four times larger because of the four men. 2. this can be simplified to Beta = 20db+ 10log (4), using the properties of logarithms. 3. Solve for Beta

//Steps// 1. Recall the formula for total sound power level: Beta = 10log(I/I_o) 2. If you had N times the people shouting, I is now N times larger 3. New formula: Beta_total = 10log(N*I/I_o) 4. Simplify: Beta = 10log(I/I_o) + 10log(N) using properties of logarithms 5. Simplify further: Beta_total = Beta_single +10log(N) 6. Plug in value for N from problem 7. Done

Thank you for doing such a great job ! -- professor

__**Fifth week (26th September, 2011) --- please consider this as a review for your exam too **__


 * 1. A light string of mass per unit length 8 g/m has its ends tied to two walls separated by a distance equal to three fourths the length L of the string. A mass m **

** is suspended from the center of the string, applying a tension in the string. The acceleration of gravity is 9.8 m/s². What size mass should be suspended from the string to produce a wave speed of 60 m/s? **  GIVEN: u= .008 kg/m g= 9.8 m/s^2 v= 60 m/s tension = w/sin(theta) distance from wall to mass = 1/2L distance between walls = 3/4L v=Sqrt(tension/u)

STEPS 1. First solve for theta, this can be done by using trigonometry. 2. After theta is found you can plug that into an equation for tension, and plug that into another equation to solve for m

v=Sqrt( m g/sin(theta)/u) 3. Solve for m


 * 2. A simple pendulum consists of a ball of mass M = 5 kg hanging from a uniform string of mass m = 0.6 g and unknown length L. The pendulum is located in gravitational field g = 9.8 m/s² and has period T = 2 s. **

** Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. (For **** simplicity, neglect the string’s weight compared to the ball’s weight. Likewise, neglect the ball’s radius compared to the string’s length). Answer in units of m/s. **  **GIVEN** mass= 5 kg T= 2 seconds g= 9.8 m/s^2 T=2pi*sqrt(L/g) v=sqrt(tension/u)

STEPS 1.Use the equation T=2pi*sqrt(L/g) to solve for the length. 2.Divide the given mass by the length to get u. 3.Plug the given values into the equation v=sqrt(tension/u) to get the velocity __ **Fourth week (19th September, 2011)** __

01. A transverse wave of frequency 35 Hz propagates down a string. Two points 5 cm apart are out of phase by π/3

GIVEN: f=1/T V=lambda/T V=f lambda V=distance/time

i. What is the wave length of the wave? (Hint : one full wavelength is 2π apart !)

STEPS 1. Convert 5cm to meters 2. If pi/3 is 5cm then 2pi must be 6 times 5 cm because pi/3 is one sixth of 2pi. 3. the answer is the wave length (lambda)

ii. At a given point, what is the phase difference (as a multiple of π ) between two ‘displacements for times 5 ms apart?

STEPS 1. Recall the equation V= f * lambda, and using the wavelength you solved for in the previous step and the given frequency, solve for v 2. use the velocity to figure out how far the wave travels in 5ms 3.divide that number by the distance for one full wavelength 4. multiply that decimal by 2pi that is the phase difference in multiples of pi

02. At t = 0, a transverse wave pulse in a wire is described by the function y (x, t= 0) = 6/( x²+ 3), where x and y are in meters. The pulse is traveling in the positive x direction with a speed of 4.5 m/s. What is y at time = t? (Hint : find y(x,t) ). STEPS 1. take the derivative of the given equation and then and set that equal to the given velocity 2. use that to solve for x 3. plug the x back into the first equation to solve for (y(x,t)) 4. use that to solve for a the coefficient that is multiplied by in the origInal equation 5 use those pieces to put the equation together.

__**Third week**__
 * 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? **

KE=1/2 * m * V^2 U(t)=1/2 * k * X(m)^2 KE=U(t) Mass=100kg k=5x10^6 V(final)=0 Xm=.0316
 * GIVEN **

**STEPS** 1. Then energy before the impact and after the impact have to be equal and that all of the kinetic energy becomes potential energy by way of the spring. 2. Set the equation for potential and kinetic energies equal to each other and solve for V using the other given values.

1/2 * m * V ^2=1/2 * k * X(m)^2


 * 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. **

**GIVEN**

**Vmax=60cm/s**
**Vmax=Xm*omega** **omega-sqrt(k/m)**

**STEPS** **i. calculate the mass** **1. Use Vmax=Xm*omega and solve for omega** **2.Use that in mass=k/omega^2** **3.Solve For mass**

ii. Find the period of the motion.

T=2pi/omega

iii.Calculate the maximum acceleration of the block.

a(max)=omega^2*Xm

X(m)
i. Find the maximum speed of the block.

1.Vmax=xm*omega......and omega=sqrt(k/m) 2.therefore Vmax=Xm*sqrt(k/m)

ii. Find the speed of the block when it is 4 cm from the equilibrium position.

1.Set up .4=Xm*cos(omega*t) using the X(t) equation and solve for t 2.Plug the value for t into the Velocity equation and get V at x=.4

V(t) =Xm*omega*sin(omega*t)

iii. Find its acceleration at 4 cm from the equilibrium position.

1. Do the same thing but this time with the acceleration equation

a(t) -omega^2*Xm*cos(omega*t)

__**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.**

**GIVEN** **Vcylinder=r^2*pi*Length** **m= Desity*Volume** **W=mg** **Fb=W(displaced-liqiud)**

**STEPS** **1. Solve for the mass of a single log using the dimensions to find the volume then the mass using the equation above.** **2.Add the mass of all six logs and solve for the weight of the logs.** **This will be the force acting in the negative y direction.** **3. Multiply the total volume of the logs by the density of the water, this is the mass of the liquid it will be displacing.** **4. Multiply the mass of the displaced water by g and you get Fb (the Buoyant force).** **5. Find the difference in the weight of the raft and the buoyant force then divide that by 200 to get the number of children the raft can support**


 * 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? **

**GIVEN** **Vcube=L*W*H** **m= Desity*Volume** **W=mg** **Fb=W(displaced-liqiud)**

**STEPS** **1. First solve for the volume of the wooden block, then multiply by the density to get the mass of the block.** **2. Multiply the mass of the block by g to get the force acting in the negative y direction** **3. Because we are solving for the depth the block will sink before it begins to float we need to set Fb equal to Wblock and solve for h**

**Weight Of Wood=Buoyant Force**
 * Ex:**

**(Length)(Width)(Height)(Gravity)(Density of Wood)=** ** (Length)(Width) (Height) (Gravity)(Density of Liquid) **

**4. Solve for the unknown height on the Fb side of the equation, and that height is the depth of the bottom of the cube**

//**Professor's Note**: **Great answers. Very well done. Good Job !!**//