Ashley+Keller

Wikispaces Quiz Questions 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.
 * WEEK ELEVEN: **

Given: A=95 cm3 ΔT=112°C α=5x10-6 °C -1

1.) If you assume that the rectangular plate is a square, you can find the length of each side to be the square root of the area. This gives you L 1 . 2.) In order to calculate the difference in area, you must first find the difference in length. In order to find <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">ΔL, use the formula for linear expansion: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">ΔL=(L 1 )(α) ΔT <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.)Use ΔL and L 1 to find L 2 by adding them together. (ΔL+L 1 =L 2 ) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.)To find the new area, A 2, use the new length, L 2 .: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">A 2 =L 2 2

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">5.)To find the change in Area, ΔA, use the formula: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">ΔA=A 2 -A 1

<span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">2. A quantity of ideal gas at 10 ºC and 100 kPa occupies a volume of 2.50 m³. <span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">(a). How many moles of the gas are present?

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">P= 100kPa <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">T= 10°C (convert to Kelvin) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">V=2.50m3 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">R=8.31 J/molK <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">n=?

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.)Since the problem states that this is an ideal gas, you can use the equation PV=nRT. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.)Plug in the given values making sure that you have the right units.

<span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">(b.) If the pressure is now raised to 300 kPa and the temperature is raised to 30 ºC, how much volume <span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">does the gas occupy? Assume there are no leaks.

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">P= 300kPa <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">T=30°C <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">R=8.31 J/molK <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">n=answer from part a <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">V=?

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.)The number of moles is constant as well as R. Plug in the new values of P and T along with R and the <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">n that you found in part a in order to solve for the new volume. Make sure you use the right units.


 * WEEK TEN: **

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?

Given: L=145cm λ=643nm d=0.15mm 1.)Because you are finding the distance between the bright fringes in a double slit system, use the equation: **dsinθ=mλ**  2.) Since θ is small, you can approximate **sinθ=y/L** (y is the distance between bright fringe). The formula should now look like this: **d*( y/L)=mλ.** 3.) Rearrange the formula so that you can solve for y: **y=(mλL)/d** 4.) Since the distance between each fringe is 1 integer, m=1. So the formula is **y=λL/d.** 5.)Plug in your given values but make sure they all have the same unit!

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.

Additional given information: t=.19μm n=1.2

1.)When a wave goes through a medium, you can use the equation λn=λ/n to get the new wavelength. 2.)The number of wavelengths of the light coming from the slit with no thin film can be found with the equation: r2 is the distance between the slit with no thin film to the central axis and r1 is the distance between the slit with the thin film to the central axis. 3.)The number of wavelengths of light coming from the slit with the thin film can be found with this equation: N1=(r1+t)/λ + t/(λn)= (r1+t)/λ + t/λn= **(t(n-1)+r1)/λ**.  4.)The central maximum is where N1 is equal to N2. Set the equations equal to each other: **(t(n-1)+r1)/λ=(dsin** ** θ **** +r1)/λ. ** 5.) Rearrange the equation to solve for sinθ. The equation should now look like this: 6.) **sinθ = y/L**. Plug this into the equation and rearrange so that you can solve for y. It should look like this: 7.) plug in the given information making sure that all of the units are the same in order to solve for y.
 * N2=r2/λ=(dsinθ+r1)/λ **
 * sinθ= t(n-1)/d**
 * y **** = (L*t(n-1))/d **
 * Professor's Note: ** ** Great work Ashley, please learn all the steps, Thank you for actively participating in this exercise. Good job ! **


 * WEEK NINE: **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. <span style="font-family: 'Arial','sans-serif'; font-size: 12px;">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 <span style="color: #800080; font-family: 'arial','sans-serif'; font-size: 12px;">at a distance d = 20 cm from the fly ; the image has the same orientation as the fly and a height = 2 H.

<span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 12px;">a) What are the focal length of the lens and the object distance of the fly? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">h=H <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">h1= 2H <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">-i-p=20 cm (i is a virtual image) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.) m can be determined by the equation h1/h. Since it is the same orientation, m is positive. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.)Use the equation m=-i/p to solve for i. It should look like this: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=-i/p <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">-mp=i <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.) Use the value you got for m to find the value of i in terms of p. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.)Plug in this value of i to the equation – i-p =20. So mp-p=20. Solve for p. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">5.)now solve for i using the equation from 2 and the value you got for p in 4. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">6.) Use the formula 1/f=1/i+1/p <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">*make sure that i is negative <span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 12px;"> The f ly then leaves lens 1 and sits in front of a second lens (say lens 2), which also forms an image at d = 20 cm <span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 12px;">from the fly and has the same orientation as of the fly, but now the image height is 0.5 Hb) What are the focal <span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 12px;">length of the second lens and the new object distance? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=.5H/H= 1/2 (+ because orientation is the same) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">p+i=20 cm <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.)Use the equation m=-i/p to solve for i. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.)The equation p-i=20 is instead of –i-p=20 because now –i is less than p. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Plug in the value you got for i in this equation to solve for p. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.) Solve for i using formula from 2. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.) Use formula 1/f=1/i+1/p to solve for f.

<span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">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? <span style="font-family: 'Calibri','sans-serif'; font-size: 16px;">1.) You can find the θ1 because you are given the vertical and horizontal distance of the corner from the observer. To find the angle use tan-1(L/D).

<span style="font-family: 'Calibri','sans-serif'; font-size: 16px;">2.Since the ray is along the horizontal once it reaches air, θ2=90°. Now that you know both angles and one index of refraction (for air n2=1), you can solve for the other index of refraction using snell’s law.

<span style="font-family: 'Calibri','sans-serif'; font-size: 16px;">n1 θ1=n2 θ2

<span style="font-family: 'Calibri','sans-serif'; font-size: 16px;">3.)Isolate n1 and solve using the variables you found.


 * WEEK EIGHT: not exactly sure if i interpreted this problem correctly. **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. Initially unpolarized light is sent into a system of three polarizing sheets whose polarizing directions make angles of Ø1 = Ø2 = Ø3 = 50 <span style="font-family: 'Cambria Math','serif'; font-size: 13px;">⁰ <span style="font-family: 'Arial','sans-serif'; font-size: 13px;"> with the directions of the y axis. What percentage of the initial intensity is transmitted by the system? (hint: be careful with the angles) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Ø1 = Ø2 = Ø3 = 50° <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.) Note that initially **//unpolarized//** light is sent into the system. This means that the one-half rule <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">must be used to calculate for I1. The equation looks as follows” <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">I1= ½ Io <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.) Next, the equations for I2 and I3 are made by the cosine squared rule. This rule is used only <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">when **polarized** light is sent into the system. The light was polarized by the first polarizing sheet, <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">so these equations can be used: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">I2=I1cos2(θ2-θ1) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">I3=I2cos2(θ3- θ2) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.) Since the values for θ1, θ2, and θ3 are the same, θ2- θ1=0°, and θ3- θ2=0. According to the <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">textbook, “the transmitted intensity I is a maximum and is equal to the original intensity Io when <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">the original wave is polarized parallel to the polarizing direction of the sheet (when θ=0° or 180°).” <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">So these equations can be rewritten as: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">I2=I1= ½ Io <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">I3=I2= ½ Io <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.) The percentage of the initial intensity is given by I3/Io. Take the equation for I3 and divide each <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">side by Io so that it looks like this: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">I3/Io= ½ <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">5.) Multiply this by 100 to get the percentage.


 * WEEK SEVEN: **

<span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">1.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. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">λ(light)=434 nm=4.34*10^-7 m <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">λ(observed)=462 nm=4.62*10^-7 m <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">c=2.99792*10^8 m/s <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">f(observed)=? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">f(source)=? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">v(observed)=?

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.)Use the wavelengths that are given to find frequencies using the equation: **fλ=c** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">-use λ(light) to calculate f(source) and λ(observed) to calculate f(observ ed).

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.)Since you are stationary and the galaxy is moving, you can rearrange the equation for finding the observed frequency in order to solve for the observed velocity: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.) Rearrange the formula to solve for v(observed): <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.)solve for v(observed) by plugging in your given data.
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">f(observed)=f(source)*(1/(1+v(observed)/c) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">c(f(source)/f(observed)-1)=v(observed) **

<span style="color: #800080; font-family: 'Arial','sans-serif'; font-size: 13px;">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.

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Use equation B=10log(I/Io) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.) In this problem, B1 (the decibel level) is given as 20 dB and you are trying <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">to find the intensity of four men’s voices. So: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">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: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.)Remember that when you are multiplying in logs, you can add them. So this equation becomes: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.)Since 10log(I/Io)=B1, you can replace that term of the equation with B1:
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">B1=10log(I/Io)=20 dB (given) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">B4=10log(4I/Io) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">B4=10log(I/Io)+10log(4) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">B4=B1+10log(4) **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Plug in the value given for B1 and solve for B4.

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Professor's Note : Very good answers -- good job ! <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. A light string of mass per unit length 8 g <span style="font-family: 'CMMI10','serif'; font-size: 13px;">/ <span style="font-family: 'CMR10','serif'; font-size: 13px;">m has its ends tied to two walls separated by a distance equal to three fourths the length <span style="font-family: 'CMMI10','serif'; font-size: 13px;">L <span style="font-family: 'CMR10','serif'; font-size: 13px;">of the string. A mass <span style="font-family: 'CMMI10','serif'; font-size: 13px;">m <span style="font-family: 'CMR10','serif'; font-size: 13px;">is suspended from the center of the string, applying a tension in the string. The acceleration of gravity is 9 <span style="font-family: 'CMMI10','serif'; font-size: 13px;">. <span style="font-family: 'CMR10','serif'; font-size: 13px;">8 m <span style="font-family: 'CMMI10','serif'; font-size: 13px;">/ <span style="font-family: 'CMR10','serif'; font-size: 13px;">s² <span style="font-family: 'CMR9','serif'; font-size: 13px;">. <span style="font-family: 'CMR10','serif'; font-size: 13px;">What size mass should be suspended from the string to produce a wave speed of 60 m <span style="font-family: 'CMMI10','serif'; font-size: 13px;">/ <span style="font-family: 'CMR10','serif'; font-size: 13px;">s? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Given: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">g=9.8 m/s^2 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">d(between walls)=3L/4 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">d(of string from mass to wall)=L/2 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">v=60m/s <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">u=8 g/m <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">theta=? (solved in #2)
 * WEEK SIX:NO QUIZ **
 * WEEK FIVE: **
 * <span style="color: blue; font-family: 'Calibri','sans-serif';">Fifth week (26th September, 2011) --- please consider this as a review for your exam too **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.) By creating a force diagram, you can create a formula that will help you solve the problem. Make a triangle with the T as the hypotenuse and the W as the opposite side of angle theta. From this you can create an equation: (there are 2 Tensions that are equal) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">W=2Tsin(theta) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">mg=2Tsin(theta)
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">T=mg/(2sin(theta)) **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">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:

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">(note that you include L in the equation and they will cancel out)
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Cos(theta)=adjacent/hypotenuse **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.)Now you can modify the velocity equation in order to solve for m.

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Substitute the formula you made for tension(from #1): <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">(velocity is the square root of mg over 2sin(theta)u) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Now isolate m so that you can find the mass. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">(mass is 2 times v squared times sin(theta) times u all over g.)
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">V=(T/u)^.5 **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">V=(mg/(2sin(theta)u))^.5 **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">V^2= mg/(2sin(theta)u) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">*m=(2v^2sin(theta)u)/g **

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.) Plug in all of your given values but make sure they all have the correct units.

<span style="color: #800080; font-family: 'calibri','sans-serif'; font-size: 15px;">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: M=5 kg m=.6g—convert to kg L=? G=9.8m/s^2 T=2 s u=? (#2) F=? (#3) V=?(#4) 1)The equation for period is T=2*pi(L/g)^.5. Use this to find the Length L of the pendulum.

T=2*pi(L/g)^.5 T/(2*pi)=(L/g)^.5 T^2/(4*pi^2)= L/g

(L is T squared times g all over 4 times pi squared) -Plug in known values to solve for L.
 * T^2g/(4*pi^2)=L**

2.)Find the mass per unit length using equation **u=m/L** (make sure you are using mass of string and in the correct units(kg)) 3.)Since the pendulum is hanging vertically and stationary, the tension (F)is defined as: -Plug in your known values to solve for F
 * F=Mg**
 * make sure you use the mass of the ball this time

4.)Speed is defined as **v=(F/u)^.5**. Plug in the values you found for each part to solve for v.

I did visit, commented on other pages and will visit again soon !! prof Week 4:
 * I responded to the note you made at the bottom of the page- Ashley **

<span style="color: #c00000; font-family: 'Arial','sans-serif'; font-size: 13px;">01. A transverse wave of frequency 35 Hz propagates down a string. Two points 5 cm apart are out of phase by π/3i. 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?

i. 1.) You can solve for the wavelength by creating a proportion. You know that the **distance** between the two points is 5 cm and they are **out of phase by pi/3**. You also know that 1 full wavelength is **2pi** apart. So when you set up the proportion it should look like this:

.05/(pi/3)=Lambda/(2pi)
 * Lambda = wavelength

2.) Isolate Lambda and solve: Lambda=(2pi*.05)/(pi/3)

ii) You can solve for the phase difference by creating a proportion. You know that the time **(period)** between the two points is .005s. You also know that a full **wavelength** is **2pi** and that the **period is the inverse of the frequency.**

(.005)/Phase=(1/f)/2pi

2.) Rearrange the formula to isolate and solve for phase difference: Phase=(2*pi*.005)/(1/f)

<span style="color: red; font-family: 'Arial','sans-serif'; font-size: 13px;">02. At t = 0, a transverse wave pulse in a wire is described by the function y <span style="color: red; font-family: 'CMR10','serif'; font-size: 13px;">( <span style="color: red; font-family: 'CMMI10','serif'; font-size: 13px;">x, t <span style="color: red; font-family: 'CMR10','serif'; font-size: 13px;">= 0) = <span style="color: red; font-family: 'Arial','sans-serif'; font-size: 13px;">6/( x² <span style="color: red; font-family: 'CMR10','serif'; font-size: 13px;">+ 3), <span style="color: red; font-family: 'Arial','sans-serif'; font-size: 13px;">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) ) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">You know: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">V=4.5 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">y’(x,t)=v(x,t) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">t=0 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">y’(x,t)= -12x/(x^2+3)^2=v(x,t) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">*note that only the term (x^2+3) is being squared, not the entire equation <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.) Next, integrate this equation from 0 to 4.5*t to get back to the position equation with respect to t. This should yeild a result for any t value you apply. **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">*Not positive if this is correct (could not get the derivative to equal 4.5 so i tried this way instead. I will check your page when you post the results) **

Week 3: <span style="color: red; font-family: 'Arial','sans-serif'; font-size: 13px;">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="font-family: 'Arial','sans-serif'; font-size: 13px;">** What you know: ** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.) The problem states that no energy is lost during impact with the wall. This tells you that the Kinetic and Potential Energy are equal to one another. **KE=PE** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.) The equation for **KE=1/2mv^2** and the equation for **PE=1/2k(xm)^2**. Set these equations equal to one another to get **1/2mv^2=1/2k(xm)^2.** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.) Modify this equation so that you can solve for v. You should end up with this equation: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.) Plug in your known values to solve for v.
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=1000kg **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">k=5x10^6 N/m **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">xm=.0316 m **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">v=sqrt(k(xm^2)/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. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. You can modify the equation **w=sqrt(k/m)** to solve for m. The resulting equation is **m=k/(w^2).** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2. Now you need to substitute the formula for w into the equation. Use the formula **v=wx** and rearrange it so that w is on one side. You should get **w=v/x**. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3. Put this equation into the first equation that you found. It should look like this: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">*note that only (v/x) is being squared, not the entire equation. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4. Plug in all of the known values to solve for m. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">**ii.** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">T=? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. The equation for T is **T=2pi(m/k)^.5**. Plug in the values for m and k and solve for T. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">*note that only (m/k) is raised to the ½. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">am=? <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. The equation for am is as follows**: am=w^2(xm**). Plug in the value you found for w and the given value for xm to solve for am.
 * <span style="font-family: 'Calibri','sans-serif';">i <span style="color: blue; font-family: 'calibri','sans-serif';">. **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">What you know **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">k=6.5 N/m **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Xm=.1 m **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">v=30 cm/s=.3 m/s **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">w=? (omega) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=? **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=k/(v/x)^2 **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">iii. **

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. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">**<span style="font-family: 'Arial','sans-serif';">i. ** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. Since you are finding the max speed, use the equation **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">vm=w(xm) ****<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">. ** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2. Since you don’t know the value for w, substitute the equation **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">w=(k/m)^(.5) **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;"> into w in the equation for vm. The resulting equation is as follows: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3. Plug in all of the known values to solve for vm. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. Use the equation **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">v=-xmwsin(wt) **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;"> to find the velocity when the block is .04 meters from the equilibrium point. First you need to find t by using the formula **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">x=xmcos(wt ****<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">). ** <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2. Once you have solved for t, you can plug the value into the first equation. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. Use the equation **a=-w^2(xm)cos(wt)** with the same variables and the answer you used for the previous question.
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">What you know: **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">m=.4 kg **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">k= 12 N/m **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">xm=.08m **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">w=? (omega) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">vm=? (max velocity) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">vm=(k/m)^.5(Xm) **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">* **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">note that (k/m)^.5 is the square root of (k/m)
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">ii. **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">x=.04 **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">w= found in i. **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">t=? **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">v=? **
 * <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">iii. **


 * <span style="color: #800000; font-family: 'Arial','sans-serif'; font-size: 13px;">Professor's Note: Great work, note the fact that it is always easy to find the term "wt". **
 * <span style="color: #800000; font-family: 'Arial','sans-serif'; font-size: 13px;">Continue to carry on the good work, this sure will develop your critical thinking abiities !! Good job **

**<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1. **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;"> 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.
 * Week 2:**

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">Known: <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">d=0.25 m so r(logs)=.125 m <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">h(logs)=1.9m (length of log) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">W(kids)=200N <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">p(logs)= 800 kg/m^3 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">p(water)=1024 kg/m^3 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">g=9.8 m/s^2 <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">FB= bouyant force

<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">1.) Use the formula V(logs)=6(r^2pi)h(logs) to find the volume all six logs. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2.) Use the formula V(logs)p(logs)g=W(logs). <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">* you have the values for V(logs), p(logs) and g, so you are solving for W(logs) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">3.) Use the formula Vp(water)g=FB to find the buoyant force. <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">4.) Since FB is the max weight that the logs can hold without sinking, use the formula FB-W(logs)=W(support) ( this represents how much weight the logs can support.) <span style="font-family: 'Arial','sans-serif'; font-size: 13px;">5.) Since you are trying to find how many kids can be supported on the raft and you know weight of each child, divide W(support) by W(kids) to find the number of kids that the logs can support.

**<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">2. **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">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 1000 kg /m3, that of liquid is 1240 kg/m3. How far below the liquid surface is the bottom face of the cube?

<span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">known: <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">s(wood)=0.015m <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">V(wood)=s^3 <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">p(wood)= 1000 kg/m^3 <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">p(liquid)= 1240 kg/m^3 <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">A(object)=s^2 <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">FB=bouyant force <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">g= 9.8 m/s^2 <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">h=?

<span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">1.) Use the formula W(wood)=FB

<span style="color: black; font-family: 'arial','sans-serif'; font-size: 13px;">p(wood)gV(wood)=p(liquid)gAh <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">Solve this equation for h. **<span style="font-family: 'Arial','sans-serif'; font-size: 13px;">(and also i realized that i did number two wrong for week 3 but I visited your page to see the correct way. **
 * <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">Professor's Note : Good Job Ashley,, keep up the good work, have I missed commenting on the page or did you erase the comment? **
 * <span style="color: black; font-family: 'Arial','sans-serif'; font-size: 13px;">- I don't think you commented, but i saw your note at the top- Ashley **