Professor's+page+-+problems+and+help

__**First week (Fall 2011**__) -- Quiz 1 Pay attention to the material covered in the class.
 * Density and Pressure ( Chapter 14 section 3)
 * Fluids at rest (Chapter 14 section 4)

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 * An interesting video on Buoyancy : **

__**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.


 * Steps to solve the problem:**

**Step 1** (from Erik's page)
 * The first step to solving this problem (and all others for that matter) is drawing a picture and then writing down your givens, which in this case you are already given a log diameter and length, the weight a child, and the density of both the logs and sea water. For this problem it can be assumed that the gravity is that of earths (9.81 m/s^2) **


 * Step 2 (from Karen -Tianelli's pages) **

F(bouyant) - W(log) =W(of the children which is unkown)
 * the sum of the force in the y direction: **
 * F(sum) = - F(bouyant) + W(log) +W(of the children which is unkown) **
 * 0 = - F(bouyant) + W(log) +W(of the children which is unkown) **

**Step 3** (from Cole - ABroatch- Rajin's pages)

Vcylinder = r²*(pi)*Length m = Desity*Volume W = mg Fb = W(displaced fluid)

Note : use the radius when finding the V and not the diameter
 * 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). **
 * Fb = V*p(liquid)*g **
 * 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?

**1**. **The equation for this problem is (volume of liquid)(density liquid)(gravity) = (volume of cube)(density of cube)(gravity).**
 * Steps to solve ( ** From Mike -Cordone's page)
 * 2. For the liquid side of the equation, you will find the volume by (length)(width)(height). but the height will stay as h, because that is what we are solving for. **
 * 3. The right side of the equation, all of the information is given to complete. the volume is (length)(width)(height), the density is given, and gravity is 9.81. **
 * 4. So now just solve for h, and this is your answer. the answer will be less than 1.5 cm because this is how tall the cube is. **

**Third week (12th September, 2011)**

**Simple Harmonic motion**

media type="youtube" key="SZ541Luq4nE" height="278" width="358" align="center" ]**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? **<span style="color:

(**From Erik's page)**
 * Step 1 **
 * The first step to solving this problem is to write down all the givens (m=1000 kg, K=5*10^6, x=3.16 cm) and then draw a picture (supplied by the professor on the professor's page). **


 * Step 2 **
 * Since this is an energy problem, solving it actually becomes quite simple. The equation for kinetic energy is K.E.=1/2m*v^2, and since the problem states that energy is conserved, the potential energy of the car should be equal to the kinetic energy of the car. The potential energy can be found using the equation P.E.=1/2K*x^2. **


 * Step 3 **
 * Since the problem states that the energy is conserved, the latter two equations given in the previous step should be equal to each other. From there the only variable that is unknown is that of v, which can be solved algebraically by plugging in your givens! **

i. Calculate the mass of the block.
 * 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.

ii. ** Find the period of the motion. **
 * T = 1/f **
 * f = angular freq. (w)/2pi **
 * iii.Calculate the maximum acceleration of the block. **
 * a (max) = w^2x(max) **

i. Find the maximum speed of the block.
 * 3.** A 0.4 kg block attached to a spring of force constant 12 N/m oscillates with an amplitude of 8 cm.

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

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


 * __From Eli's page__ **


 * Step 1 **

More givens and more pictures. To find the maximum speed of the block I would use multiple equations. First, I would find omega with the equation omega=sqrt(K/m). Then I'd use the equation Vmax=Xmax * omega
 * Step 2 **


 * Step 3 **

To find the speed of the block at 4 cm I would first use the equation X(t)=Xmax * cos(omega * t) to solve for t. Then, I would use the V(t)=-Xmax* omega *sin(omega*t) to find the velocity at .04 m by plugging in the t found in the first equation. Finally, to find the acceleration at .04m I'd use the equation A(t)=-omega^2 * X(t)
 * Step 4 **

**Fourth week (19th September, 2011)**

**Enjoy resonance :**

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

i. What is the wave length of the wave? (Hint : one full wavelength is 2π apart !) ii. At a given point, what is the phase difference (as a multiple of π ) between two ‘displacements for times 5 ms apart?


 * Steps to solve: (taking from Adam, Sam, Amanda and Rajin pages) **


 * //Steps// **
 * //One full cycle = lambda wavelength = T period = 2// π phase.**
 * 1. Set up proportion: (distance between points)/(pi/3) = (wavelength)/(2pi) **
 * 2. Solve for wavelength **

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. Find the period using T = 1/f **
 * 2. Remember that one oscillation covers a phase of 0 to 2pi **
 * 3. Set up a proportion: (phase difference/0.005s)=(2pi/period) **
 * 4. Solve for phase difference **
 * 5. Divide by pi to get your answer in terms of pi radians --> This is an important step **

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 di rection with a speed of 4. 5 m / s. What is y at time = t? (Hint : find y(x,t) ) **Steps:** **1. dx/dt = 4.5** **2. Integrate dx = integrate (4.5 dt)**
 * 3. Use limits from x to x1 and time from 0 to t**

**4. Find x1 in terms of x and t** **5. Substitute the new x (or x1 ) in the y (x,t) equation** -- **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? Steps to solve the problem:
 * From Eli, Ashley and Rajin's pages**
 * Step 1 .** begin by drawing a picture and stating my givens.
 * Step 2 .** The force of tension in the y direction for either half of the string is equivalent to T * sin(theta), so the tension in the y direction holding the mass is 2T*sin(theta). So mg=2T*sin(theta) => T= mg/**2**sin(theta) ---> this is absolutely important !
 * Step 3 .** To find theta it would just be trig so cos(theta)=((3L) / 4) * 0.5) / (L/2) => cos(theta)= 3/4 => theta= 41.4
 * Step 4 **. I would then take the equation v=sqrt(T/u) and substitute for T =>v=sqrt((mg)/(2*u*sin(theta)))Because we have u, theta, g, and the desired v, all one has to do is solve for m.

2.) You can find the angle (theta) required with another triangle. This time the **hypotenuse is L/2** (since the string is in two portions) and the **adjacent side** of angle theta **is 3L/8**. This number is from dividing the overall length between the two walls by two since only half of this is the adjacent side of the triangle you are making. Use the equation:
 * Ashley's page : hint to find the angle **

(note that you include L in the equation and they will cancel out)
 * Cos(theta)=adjacent/hypotenuse**

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.


 * Tom - Wolfgang - Austen's pages **

__Solution;__ 1) Use equation T=2*pi*(L/g)^.5 2) Rework the problem to solve for the unknown L. (Should look like L=(T^2*g)/(4*pi^2)) 3) To find u the equation is simply µ = m/L 4) Since the pendulum is non-moving, the definition of tension is F=Mg 5) To solve for the velocity plug in all your found data into v=(F/µ)^.5  __**Week 6 (03rd October 2011**)__    //Exam week --- no Friday Quiz//

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. Solution 1. The first part to solving this problem is to find the frequency of both of the wavelengths. --> This is easily calculated because we know that frequency is found by calculating the velocity divided by the wavelength both of which are given to us, but the wavelength is in nanometers so we must convert that to meters first. 2. The wavelength seen in the lab will be the value for the frequency of the object and the frequency of the medium will be the know wavelength of the "red shift". 3. Now that the frequencies have been calculated we can move closer to finding how fast the galaxy is moving. 4. To do so you must use the equation for a source moving away from an observer because "red shift" is when the light is moving away from earth. 5. Once you set up the equation with all the given and calculated values all that is left to do is solve for the velocity variable above the speed of light using arithmetic

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: loudness of one man: 20 dB //I//o(intensity knot): 1X10^-12 w/m^2 //I//(intensity):? loudness of four men:?

1. Use the equation loudness=10*log(//I/I//o) to solve for //I//. 2. Once //I// is found, the plug that into the equation multiplied by four, because now there are four men, and solve for the loudness. 3. Equation will like like this; loudness=10*log(4*//I/////Io),// then just solve for loudness.


 * Further, (more complete version) **

Use equation B=10log(I/Io)

1.) In this problem, B1 (the decibel level) is given as 20 dB and you are trying to find the intensity of four men’s voices. So:

**B1=10log(I/Io)=20 dB (given)** 2.)The intensity of four men is 4 times greater, so the intensity is multiplied by four. The initial equation now looks like this with that modification:

**B4=10log(4I/Io)** 3.)Remember that when you are multiplying in logs, you can add them. So this equation becomes:

**B4=10log(I/Io)+10log(4)**

4.)Since 10log(I/Io)=B1, you can replace that term of the equation with B1:

**B4=B1+10log(4)**

Plug in the value given for B1 and solve for B4.

__** 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) I thought I posted the figure, but it has not uploaded to the site. I will post a propper solution with a figure soon --- prof

__** 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?

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? Until I post my steps to solve this proble visit Stacey's page at "sdufrane" she has set up the distance equations right --

a) ** Definitions of the terms : **

p = object distance i = image distance d = distance between image and object f = focal length M = magnification

Same orientation image --> virtual image
 * Useful information : **

virtual image --> positive M

__** Image is twice as the object --> lens type is converging or convex **__


 * ==> Image is formed on the same side of the object, and behind the object. **
 * Equations : **

i - p = d - (equation 1) Note : distances are involved, it is safe to use the absolute values M = - i /p, Mp = -i, absolute value of i = Mp Mp -p = d  [when given values are substituted (d =20 and M = 2), we get p = 20 and i = 40 ]
 * __first step__ **

__**second step**__

Thin lens equation [1/p + 1/i = 1/f]

__sign rules__

p is positive = object is in front of the lens i is negative = image is on the same side of the object (virtual image !) f is positive = converging lens [substitute the values, you get ,1/20 - 1/40 =1/f, f = + 40 cm]

(b)

** Useful information : ** Same orientation image --> virtual image virtual image --> positive M

__** Image is half as the object --> lens type is diverging or concave **__

** ==> Image is formed on the same side of the object, and bin front of the object. **

** Equations : **

** __first step__ ** p -i = d - (equation 1) Note : distances are involved, it is safe to use the absolute values M = - i /p, Mp = -i, absolute value of i = Mp p - Mp = d  [when given values are substituted (d =20 and M = 0.5), we get p = 40 and i = 20 ]

** __Second step__ **

Thin lens equation Sign rules p is positive i is negative f should come out as negative [When values are substituted, f = - 40 cm] //Learn to draw the ray-diagrams for each case//

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?

This problem was discussed in the class !

AN INTERESTING LINK, FASCINATING PHYSICS

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__ **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.

Until I post my solutions (steps) go to Ashley Keller and ABroatch pages.

Karen provides the complete answer (with diagrams and cross-referencing the textbook) and please visit her page at http://phy113-fall2011-wikis-jitt.wikispaces.com/karenB
 * My solution : **

--- __** 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.

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.

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.

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.

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?

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?