Erik+Quitzau

Erik Quitzau's Wiki Solutions

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.
 * SECOND WEEK**

Step 1 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 The next step is to write down the necessary equations. For this problem the necessary equations are as follows: Fb=Weight ( of the displaced fluid) Weight=mass*gravity Density(p)=Mass/Volume Fb,max=Volume,max*p,fluid*gravity

Step 3 This next step involves calculating the volume of the raft. To do so one must realize that volume=pi*r^2*length. Since we already have the diameter of a log, divide that by 2 an plug in your information to get the volume of one log. The total volume of the raft is then found by multiplying your previous answer by 6 as there are 6 logs.

Step 4 This step involves mass and weight of the raft. Using the density equation given above, plug in the density of the raft (density of the log) and the total volume of the raft. With this, you should have enough information to find the mass using algebraic rearrangement of the density equation given above. To find the weight, simply multiply the mass of the raft by 9.81 (gravity).

Step 5 This step involves finding the Maximum buoyant force of the salt water pushing back into the raft. Using the equation above to find Fb,max, one should have enough information from his or her givens as well as information previously found to calculate the maximum buoyant force.

Step 6 Now, as the equation above states, Fb=Weight (of the displaced fluid), so one must simply set the maximum force found in step 5 equal to the weight of the raft with the addition of the weight of x amount of children. Subtract the weight of the raft from the Maximum buoyant force and then divide by the weight of one child to find x. Remember to round down to the nearest whole number because it is not possible to have a fraction 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 Simply draw a picture and restate the givens from the problem (the cubes length, width, and heigh, the density of the wood, and the density of the liquid).

Step 2 State the equations being used in the problem; Density(p)=m/v Fb=Weight Fb=h*area*p,liquid*gravity

Step 3 Find the weight of the block by first calculating out the mass using the density equation given above. Since the block is a cube, one can calculate the volume by cubing the given side length. To find its mass simply multiply the volume found by the density of the wood. Then to find the cubes weight multiply the mass by 9.81 (gravity).

Step 4 Because Fb is equal to the weight OR h*area*p,fluid*g, one can rewrite the equation as h*area*p,fluid*g = weight. Since we already have the weight calculated, simply plug in the givens on the left side of the equation (to find area, square the given side of the block). After that, divide the weight by the product of the area*p,fluid*gravity. AMAZING! All that is left is h=your answer! (Make sure it is in meters though!) //Professor's Note: **Very good answers, (note : Fb = weigt of the displaced fluid). Everything is explained very well. Keep it up !**//

**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?

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!


 * Oh! DON'T forget to convert your x in cm to m!**

**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.

ii. Find the period of the motion.

iii. Calculate the maximum acceleration of the block.

Step 1

The first step to solving this problem is to write down all the givens (K=6.5, x=10 cm, v=30 cm/s at x= 5cm) and then draw a picture. Convert all values with centimeters into meters.

Step 2 To solve part i, use the conservation of energy equations explained in problem 1 (KE=PE; 1/2m*v^2 = 1/2K*x^2) and then simply plug in all your applicable givens. All thats left is to solve for m !!!!

Step 3 To find the period of the system use the equation Period(T) = 2pi*SquareRoot(m/K). Then using the mass you solved for in step i, plug m and K into the equation and solve to get T!

Step 4 As we all know, Force is equal to Mass multiplied by acceleration (F=MA). Similarly, Force is also equal to the negative spring constant multiplied by the amplitude (F=-Kx). Since the amplitude is equal in both the negative and positive directions of its motion plug in the negative amplitude so the negative spring constant and the amplitude equal a positive force. Thus, since F=MA=-Kx, The only variable that needs to be solved for is a. Plug in your givens and Solve for your answer!

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

//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.//

Step 1

The first step to solving this problem is to write down all the givens (K=12, xm=8 cm, x1=4 cm, m=0.4 kg) and then draw a picture. Convert all values with centimeters into meters.

Step 2 Like the last problem, use the conservation of energy equations given (KE=PE; 1/2m*v^2 = 1/2K*x^2) and plug in all your applicable givens and then solve for v to get the maximum velocity of the mass in the system. (Note: you must use the xm (maximum amplitude) for this equation)

Step 3 Again, use the conservation of energy equations given (KE=PE; 1/2m*v^2 = 1/2K*x^2) and plug in all your applicable givens and then solve for v to get the maximum velocity of the mass in the system. (Note: you must use the x1 (partial amplitude) for this equation)

Step 4 Also, like problem 2, use the equations of force set equal to each other (F=MA=-Kx) and then plug in your applicable givens and then simply solve for a. (Note: you must use the x1 (partial amplitude) for this equation)

Congratulations! you have just completed this weeks wiki!

**FOURTH WEEK**
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?

__Step 1__ To solve this week’s wiki, first write down your givens and draw a picture.

__Step 2__ To solve part i) one must understand that Φ (phase) is equal to 2 * pi * the change in distance divided by the wavelength. Through algebra, it is possible to rearrange the equation so that it looks like the following: (wavelength) = 2* pi * Change in distance / phase Since we have all but one of the above variables, simply plug in your givens to find your wavelength (Make sure to pay attention to units).

__Step 3__ To solve part ii) know that a waves velocity is equal to its wavelength multiplied by its frequency. Using the formula Φ = 2pi*velocity*time/wavelengh, one can substitute the formula v=frequency*wavelength for velocity. The values for the wavelength should cancel out and you should be left with Φ = f*t where f=frequency and t=time. Simply plug and chug for your answer but watch those units!


 * Professor's Note : As always everything is explained very well. Keep it up ! thanks for doing such a great job in explaining. **

__**Fifth Week**__

=
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? ======

Step 1 Draw a picture and state your givens.

Step 2 You need to know the angle in the corners of your picture thus the angle can be found by taking the inverse cosine of ((3/8)L/(L/2))

Step 4 Now that you know the angle find the tension using the formula 2Tsin(theta)=mg Note that you will not come up with a number but instead a formula with the only variable being that of mass. This formula is your tension.

Step 5 Now using the formula V=sqrt(Tension/u). Plug in all your givens and the formula you got for tension and then solve for your mass.

=
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. ======

__Step 1 __ Draw a picture and write down your givens.

__Step 2 __ This step helps you find your mass per unit length (u). To find u you must first know the mass of the string (given) and the length of the string (not given). The length can be found using the equation Period = 2pi*Sqrt(L/g). Since you know your period and since g is known to be 9.8 m/s^2, plug in your givens to find L. You were given the mass of the string in the problem so use your L to find u. u = mass/L

Step 3 This step helps you find the tension in the string. For this part tension will be equal to the force of gravity on the ball. Thus F = Mball * 9.8

Step 4 Its finally time to solve for your velocity. The velocity equation states that V = Sqrt(F/u). Since we already found F and u in the previous steps, simply plug in to find your velocity!

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.

First you need to find the frequencies of each wave. Because you already know the wavelengths of each and the velocity of light, use the equation (frequency) = Velocity / Wavelength. Using the frequencies you just previously found Plug in to the equation (Observed frequency) = (actual Frequency)[speed of light / (Speed of light + Vs)]. Simply 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.

Using the equation B=10log(I/Io) Where: B=20 Use simple algebra to calculate the value of (I/Io). This should be B/10 = (I/Io) Now that you know what I/Io equals, Multiply that number by 4 and plug it back into your log equation which should now be:

10log(4*(I/Io)) = your answer!!!!!!

= Week Eight =

D efine polarization

Polarization is the orientation of oscillations in the plane perpendicular to a transverse wave's direction of travel.

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)

This problem is actually quite simple once a picture has been drawn and you understand your givens and equations. Referring to Chapter 33 section 33-7, we can find a picture that ALMOST represents what we are attempting to accomplish with this problem. Looking at figure 33-13, you can see a diagonal line that lances through two parallel circles (Polarizing sheets) that are perpendicular to the line that pierces them. The problem we are trying to solve however, has 3 polarizing sheets instead of 2; so you must simply add one more of these polarizing sheets to diagram 33-14.

The next part of this problem requires an understanding of your givens. First off, it was given that each circle was rotated around the line that pierces them exactly 50 degrees from the y axis. In other words, if you picture a circle with its center point drawn (Center point will be the line piercing the circle) and a line drawn directly upwards in the y direction, then rotate the circle around the center point clockwise until the line drawn from the center point is exactly 50 degrees from the position where it initially started. All three polarizing sheets (Circles) are oriented this way.

Now to solve the problem. An equation was given in the book which states I = Io*cos^2(50). Since there are 3 polarizing sheets you can set up a set of equations that looks like: I1 = Io*cos^2(50) I2 = I1*cos^2(50) I3 = I2*cos^2(50)

So to solve this simply continuously solve the system of equations:

I1 = Io*cos^2(50) <span style="font-family: 'Times New Roman',Times,serif;">I2 = I1*cos^2(50) = (Io*cos^2(50))* cos^2(50) <span style="font-family: 'Times New Roman',Times,serif;">I3 = I2*cos^2(50) = ((Io*cos^2(50))* cos^2(50))*cos^2(50) <span style="font-family: 'Times New Roman',Times,serif;">OR <span style="font-family: 'Times New Roman',Times,serif;">I3 = (cos2(50))3 * Io

<span style="font-family: 'Times New Roman',Times,serif;">The percentage is the coefficient in front of Io multiplied by 100.

<span style="color: #800000; font-family: Tahoma,Geneva,sans-serif;">Professor's Note: good Answer. Explanation alone would be enough. keep up the good work !

=Week Eleven= 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.

<span style="font-family: 'Times New Roman',Times,serif;">Begin this problem by drawing a picture and then defining your givens, which in this case are the area of the rectangular plate, the coefficient of linear expansion, and a change in temperature. <span style="font-family: 'Times New Roman',Times,serif;"> Assume the plate is a perfect square so that you may be able to derive one of the lengths of its sides by square rooting the area (Don’t forget to convert the area to meters squared before you do this!). Now that that is done, you now have all the keys needed to solving this problem. <span style="font-family: 'Times New Roman',Times,serif;"> The equation that should be used to solve this problem is: <span style="font-family: 'Times New Roman',Times,serif;">The change in L = L* linear coefficient * the change in T <span style="font-family: 'Times New Roman',Times,serif;">Using the above equation, plug in your givens to find the change in length of one of the sides of the plate. Add that number to the original side length and the sum of the two numbers to find your new area. <span style="font-family: 'Times New Roman',Times,serif;"> Finally subtract the old area given from the new area recently found to get your answer!

<span style="font-family: 'Times New Roman',Times,serif;">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.

<span style="color: #000000; font-family: 'Times New Roman',Times,serif;">This problem does not really have a helpful picture that can be drawn to assist in solving it, so for the first step, simply state your givens of pressure, temperature, and volume. <span style="color: #000000; font-family: 'Times New Roman',Times,serif;">Note: Make sure all temperatures are converted into Kelvin and all pressures are converted into Pascals!

<span style="font-family: 'Times New Roman',Times,serif;">a) To solve part a, the equation that should be used is: <span style="font-family: 'Times New Roman',Times,serif;">For this equation, R is the universal gas constant of 8.31 J/(mol*k) <span style="font-family: 'Times New Roman',Times,serif;">Since we are given everything but the number of moles in this part of the problem, simply plug in all of your givens and solve for n!
 * <span style="font-family: 'Times New Roman',Times,serif;"> P*V = n*R*T **

<span style="font-family: 'Times New Roman',Times,serif;">b) Part b of this problem is almost exactly the same as part a. The difference between the two, however, is that there are new values for pressure and temperature, with the volume of the gas being the unknown in this equation. <span style="font-family: 'Times New Roman',Times,serif;"> After converting the pressures and temperatures into SI units, use the equation: <span style="font-family: 'Times New Roman',Times,serif;">When solving this problem, use the amount of moles found in part a, as well as the universal gas constant that was previously stated. Finally, plug in all your givens and solve for V!
 * <span style="font-family: 'Times New Roman',Times,serif;">P*V = n*R*T **


 * <span style="color: #800000; font-family: 'Times New Roman',Times,serif;">Professor's Note: Thank you Erik, for a wonderful semester. You did very well !! **