mmerk

Week 1
 * __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. a. Solve for the volume of an individual log. (pi * r2 * L) b. Multiply the volume of an individual log by six for the total raft c. Multiply the total volume by the density of the logs and then the acceleration due to gravity, for the logs weight d. Find the max buoyant force, ( Volume of Fluid * density * acceleration due to gravity) e. Set the weight of the logs (-), max buoyant force (+), and children’s weight (multiplied by n, variable of number of children) equal to zero f. round answer down to nearest whole number

**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? a. Find the volume of wood cube, (length^3) b. Multiply the volume of the wood cube by the density and acceleration due to gravity to find its weight c. Find the buoyant force by multiplying the acceleration due to gravity, the liquids density, and the area of the cube by an unknown height (h) d. By setting the weight of the block equal to the buoyant force the only unknown is height (h) so it can be solved for with algebra

Professor's Note: Matt, I missed the page at first. Great job in explaining the steps

<span style="color: #000000; font-family: 'Times New Roman',Times,serif;">//**Week 2**//

//**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? **//

a. First list all givens m=1000kg k (spring constant)= 5 * 10^6 Delta x= .0316 m final velocity= 0 m/s b. Given the spring constant and the mass the omega value can be found (k/m)^(1/2) c. With the found omega the initial position oscillation equation can be used to be put in terms of t x(t) =x //m// * cos( ω*t) using .0316 for the displaced distance at the time and the value of amplitude d. x(t) in terms of time can be plugged into the previously derived acceleration oscillation equation to solve for a numeric value a(t) = -ω^2 * x//(t)// e. Now with a numeric value for acceleration this can be plugged into a motion equation v(final)^2 -v(initial)^2 = 2* a* (x-x(initial)) with v(final) as zero, and the known distance traveled and acceleration found the only piece missing it the initial velocity which can now be solved for

//**<span style="font-family: 'times new roman',times,serif; font-size: 12px;">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. **// //**<span style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">i. Calculate the mass of the block. **// a. Use base equation ω=(k/m)^(1/2) b. substitue mass k=m *ω^2 c. solve for mass //**<span style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">ii. Find the period of the motion. **// a. With found mass use T = 2* pi *(m/k)^(1/2)

//**<span style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">iii.Calculate the maximum acceleration of the block. **// a. Max acceleration equation a(max)= ω^2* x(m) b. substitute omega and amplitude in, solve

<span style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">**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: 'Times New Roman',Times,serif; font-size: 12px;">i. Find the maximum speed of the block. **// a. First list all the defined givens m=.4kg k= 12N/m x m = .08m

b. With the spring constant and mass find omega ω= (12/.4)^(1/2) c. Plug into max velocity equation V(max)= X(max) *ω //**<span style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">ii. Find the speed of the block when it is 4 cm from the equilibrium position. **// a. Use position and solved omega and max amplitude in position equation to find time x(t) =x //m// * cos( ω*t) b. With found time and max velocity the velocity at .04 m can be foud v(t) = -v(t) sin (ωt) --- check this equation you are missing vm //**<span style="font-family: 'Times New Roman',Times,serif; font-size: 12px;">iii. Find its acceleration at 4 cm from the equilibrium position. **// a. With found time time into the acceleration equation and enough variables are solved for a(t) = -ω^2 * x//(t)//


 * //Week 4//**
 * 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 !)**

<span style="font-family: Arial,sans-serif; font-size: 10pt;">a. phase constant = distance separated * (2 π <span style="font-family: Arial,sans-serif; font-size: 10pt;"> / wavelength) <span style="font-family: Arial,sans-serif; font-size: 10pt;">b. by multiplying both side by wavelength and dividing both by the phase constant the equation yields <span style="font-family: Arial,sans-serif; font-size: 10pt;">wavelength = distance separated * (2 π <span style="font-family: Arial,sans-serif; font-size: 10pt;"> / phase constant) <span style="font-family: Arial,sans-serif; font-size: 10pt;">c. with distance between the points at .05 m and the phase constant of π <span style="font-family: Arial,sans-serif; font-size: 10pt;">/3 wavelength can be calculated


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

<span style="font-family: Arial,sans-serif; font-size: 10pt;">a. convert 1ms into seconds by dividing it by 1000 <span style="font-family: Arial,sans-serif; font-size: 10pt;">b. use the given frequency to find the period <span style="font-family: Arial,sans-serif;">I got stuck on this one Okay, here is some help, for one period (T), the phase is 2π radians Use the information from above to find T phase diff = 2π (5 ms/ T)

**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) )** <span style="font-family: Arial,sans-serif; font-size: 10pt;">a. Take the derivative of y(x,t) with respect to x as that is the displacement from the x axis, this is a partial derivative but because there is no t value it works out as a standard derivative <span style="font-family: Arial,sans-serif; font-size: 10pt;">b. y’(x,t) = 4.5 which should be set to the found derivative to solve for x <span style="font-family: Arial,sans-serif; font-size: 10pt;">c. with a solved value of x plug it back into the original equation to find y (x,t=0)
 * 02. At t = 0, a transverse wave pulse in a wire is described by the function**

<span style="font-family: Arial,sans-serif; font-size: 10pt;">** Professor's note ** : you may need an integration to find the new x (after a time t) and then plug back in the equation

//**WEEK 5**//

**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?** a. First draw a picture so it is easier to see the quantities given and the problem missing values

b. Convert given mass per unit to SI units, kg, my multiplying the value by 10^(-3)

c. Set up a triangle with the given side's lengths

d. Solve for the angle theta using trig, sin^(-1) ((3/8)/(1/2))

e. Now with a value of theta the velocity along a stretched string can be found

f. with given velocity, μ, the angle, and gravity the unknown mass can be solved for

g. remember that the mu value must me multiplied by 2 because there are two lengths of string

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">**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.**

a. Once again draw a diagram to see the problem easier

b. Convert the g given into kg

c. With the given period the simple pendulum equation T= 2 (pi) (L/g)^(1/2) can be used

d. The only missing value is L and can be solved for with algebra

e. Now with solved L mu can be solved for by dividing the kg of the string by the found L

f. Now with the equation for the velocity in a stretched string V can be found

g. Remember to convert the mass of the mass into N by multiplying by the given acceleration due to gravity

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. a. Convert nm to m (1nm = 1 x 10^(-6)m) b. With the speed of sound and the wavelengths convert to get frequency (speed of light = wavelength * f) c. Use Doppler's equation for source moving away, with the speed of light being used instead of the speed of sound
 * Week 7**

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.

Professor's Note : Good job on the first problem and please refer to my page for the second !


 * Week 7**

Define polarization Waves with uniform orientation perpendicularly to the axis 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)

a. Use picture of problem 34 in book to give a good visual of the problem b. For the first sheet use I = .5 * Io, this can be used because the problem states that the incoming light is unpolarized c. For the second sheet the I =Io cos^2(theta) will be used because the light coming in is polarized. Theta is equal to 80 which is the angle between the two different directions of the lenses. d. For the third sheet the equation I =Io cos^2(theta) is used again because the light is still polarized, but theta is equal to 100 this time. e. setting all three equations equal to each other forms I3 = .5 Io *cos^2(80) *cos^2(100)

I3/ Io = 4.55* 10^(-4)

Therefore about .0455% of the initial intensity emerges

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.
 * Week 8**

a) What are the focal length of the lens and the object distance of the fly?

a.Draw the picture so there is a physical outline of the problem --> draw the picture, it always helps b.Make sure you know that the image is virtual because it is on the same side as the fly, use a negative sign for the i term --> good point c.The absolute value of magnification of the fly is two and plug into the formula of m = -i/p d. The m must be negative because the i term is also and the p, object distance, is postitive ---> did you mean both i and p are positive, not the case here, for virtual image i is negative e.Because the image is in distance from the fly not the lens i = p+20 --> correct f.The p term can then be solved for g. With both the p and i terms the focal length can be found using 1/p + 1/-i = 1/f --> again, you are using the correct sign for i here

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?

a. Once again draw a picture b. The i term is once again negative if it has the same orientation as the fly c. Now the absolute value of the magnification is .5 and use as negitive again in the formula m = -i/p d. Once again the image is from the fly so -i = p+20 e. Solve for p f. With both the p and i terms the focal length can be found using 1/p + 1/-i = 1/f

**Professor's Note:** Please understand that you are dealing with two different lenses here. One is converging (makes a bigger virtual image) and the other is diverging (makes a smaller virtual image). When you draw the ray diagrams, they look obvious and can set up the correct distance relationship. i - p = d or p - i = d etc.

Still a good effort -- refer to my page later.

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?

a. First find the angle that is formed by taking the inverse tangent of (L/D) making sure the angle is with the perpendicular b. Use Snell's Law n1 * sin(**Θ**) = n2 * sin(**Θ**) c. n1 is the index of refraction of air, 1, at a perpendicular angle, 90 degrees d. The second theta was found in step a, so now n2 can easily be found

__ **Week ten (14th of November, 2011)** __

<span style="background-color: #ffffff; font-family: 'Times New Roman',Times,serif; font-size: 18px;">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?

1.Draw a picture 2. lambda is given as 643* 10^(-9) m, d= .15 * 10^(-3)m, and D =1.45m 3. Use d * sin(theta) =m* lambda for bright fringes 4. Remember sin(theta) approx tan(theta) = y/D 5. For the center fringe take m = 0 6. For the first fringe from the center take m=1 7. Substitute y/D for sin(theta) in given equation 8. Know y can be solved for and because the value at m=0 this is the distance

<span style="background-color: #ffffff; font-family: 'Times New Roman',Times,serif; font-size: 18px;">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. <span style="background-color: #ffffff; font-family: 'Times New Roman',Times,serif; font-size: 18px;">As a result, the central maximum of the interference pattern moves upward a distance y, find y.

<span style="background-color: #ffffff; font-family: 'Times New Roman',Times,serif; font-size: 18px;">I am not sure on how to do this problem. thank you for your honesty ! please refer to KarenB's page.OR check the last HW But i believe the equation we need to use is along the lines of 2t =(m+1/2) * (lambda/n2) -t= .48 * 10^(-6), n -1.5, m=1 -solve for the new lambda -with the new lamnda solve d*(y/D) = m* lambda -you know D,d from the previous problem and use 1 for m again -solve for the new y value

Week 11

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. a. Draw a picture with a square plate with each side a length of the square root of .95m b. Use the equation for linear expansion, delta L = length * coefficient of linear expansion * change in temperature c. Use 112 degrees Celsius for the change in temperature and the given coefficient of linear expansion d. With the new found length square the value to find the new value e. Subtract the old area from the new 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. (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. a. Use the equation PV = NkT b. Convert the pressure to Pa, and use 5.6705*10^-8 for the Boltzmann's constant c. This will solve for the number of moles d. Divide Avogadro's number by the found number of particles, Avogadro's number is 6.023*10^23

a. Knowing the number of molecules and either the gas constant or boltzmann's constant don't remain the same b. Setting the two equations equal to each other for before the pressure and temperature changes and after yields, (PV)/T = (PV)/T c. Once again convert pressure to Pa d. The volume can then be solved for

FINAL WEEK

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.

a. Find the change in temperature from the beginning and end states b. The molar specific heat capacity at constant pressure is given as 28.8 J/mol*K c. The heat transferred can then be found by using the equation Q= n*Cp*delta T

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.

a. Convert the volume from liters to m^3 b. Convert the pressure from atm's to pascals (1.01*10^5 pa = 1 atm) c. Because the gas in an ideal gas the equation PV=nRT holds true, with a give R, P, V, and n the temperature can be solved for d. With the temperature found now the average kinetic equation can be used KEavg= 3/2 * k* T e. Use the given value for k, Boltzmann's constant

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?