Eli+Novicky+Wikispaces

__**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.

1. First, I would draw picture of what question was asking and make a list of givens. The givens in this case include the diameter of the logs, the length of the logs, the density of the logs, the weight of an average child, and the density of the liquid it is floating on.

2. I would then look at necessary equations to solve this problem. They include: Density=mass/volume Weight= mass * g = density * volume * g FBmax= Volumemax * Densityfluid * g Volume of the logs= (Pi * r^2 * l) * number of logs (6 logs)

3. Next I would find the volume of the logs by plugging the correct number into the Volume formula. After finding the Volume of the set of logs I would find the FBmax by plugging the necessary numbers into the FBmax equation. Then I would find the weight of the logs using the equation for Weight.

4.I would then take into account that the the FBmax must be larger than the sum of the weight of the logs and children together. So... FBmax=Wlogs + Wonechild*number of children which means.... (FBmax - Wlogs) / Wonechild = number of children


 * 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. Once again, I'd draw a picture portraying the problem and make a list of givens.

2. Then I'd look at my necessary equations: Density=Mass/Volume FB=h * A * Densityfluid * g FB= Weight Volume= l*w*h W=m*g= density * volume * g

3. Then, I would find the weight of the wood using the Density and Volume to find the mass and then multiply the mass by gravity to find the woods weight. Here, it is key to note that FB=Weight. Therefore, Weight= h * A * Densityfluid * g. So... h=Weight / (A * Densityfluid * g)


 * Professor's note: Very good, this is exactly what I was expecting. Keep up the good work **


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


 * 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 I'd begin this problem by making a givens list (and converting all units to SI) and drawing a picture that I could reference.

Step 2 The problem is asking for the velocity of the car right before impact and states that energy is conserved, therefore I would set the the kinetic energy equation (KE = 1/2m * v^2) equal to the potential energy equation (PE = 1/2K * x^2).

Step 3 So, 1/2m * v^2 = 1/2K * x^2 Now you just need to plug the givens in to solve for v.

**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 I'd make another givens list, picture, and convert everything to SI.

Step 2 To solve the first part of this problem I would use a series of equations. First I would take the equation Vmax= Xmax * omega and solve for omega (multiply the given V by 2 to get Vmax). Then I would take the equation omega=sqrt(K/m) and solve to find the mass.

Step 3 In order to find the period of the motion I would use the equation T=2pi/ omega. All one needs to do is plug in the mass (which you found in part i) and spring constant to find T (period).

Step 4 To calculate the maximum acceleration of the block I would use the equation Amax=omega^2 * Xmax and just solve for Amax.

**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.

More givens and more pictures.
 * Step 1**

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**

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.
 * Step 3**

Finally, to find the acceleration at .04m I'd use the equation A(t)=-omega^2 * X(t)
 * Step 4**

Professor's Note: Good work, and it is cited in my page !! way to go ..


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

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

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?

02. At t = 0, a transverse wave pulse in a wire is described by the function y (x, t= 0) = 6/( x²+ 3), where x and y are in meters. The pulse is traveling in the positive x direction with a speed of 4.5 m/s. What is y at time = t? (Hint : find y(x,t) ) **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? Step 1  I'd 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/2sin(theta) 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. 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. <span style="display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">Step 1 <span style="display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">Pictures and givens...   <span style="display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">Step 2 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.

Step One: First, I would convert each wave length to SI units.

Step Two: Find the frequencies of each wavelength using the equation f=v/lambda, where v is the speed of light and lambda is the wavelength.

Step Three: Then I would use the equation fo=fs(1/(1+vs/v)) to solve for vs, where fo is the observed frequency, fs is the source frequency, and v is the speed of light. I would use this equation because the observed wavelength on earth is greater than the actual wavelength of the light. Therefore, the observed frequency must be lower than the actual frequency which means the galaxy is moving away from Earth which is a fixed source.

02. The sound of a man shouting at the top of his lungs from a rather large distance away from your ear has loudness of only 20 decibels. What would be the decibel level of four men shouting at the top of their (equally powerful) lungs from the same distance away from you ear? Assume that there is no interference from superposed waves.

Step One: First I would set each decibel level to 10*log(x). So => db=10*log(x) for each man.

Step Two: Then I would plug all answers into the equation 10*log. So => 10*log(ans1+ans2+ans3+ans4)= final answer Professor's Note : Very good answer on the first. Check my page for the second.
 * Week eight ( 18 October, 2011)**

Define polarization- Polarization is defined as the process or phenomenon in which the waves of light or other electromagnetic radiation are restricted to certain directions of vibration, usually specified in terms of the electric field vector

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)

Step One: First, I would note that the equation for the intensity of unpolarized light going through a polarizing sheet is I**1** = (1/2) I0. An additional equation that should be noted is I**1**=I0*cos^2(theta).

Step Two: I would first use the equation I**1** = (1/2) I0 to find I**1**. To find I**2,** I would use the second equation and plug I**1** in. So => I**2**=I**1***cos^2(theta). One could also so that I**2**=(I0 /2)*cos^2(theta). Then, I**3** =I**2***cos^2(theta)= (I0 /2)*cos^2(theta)*cos^2(theta).

Step Three: The percentage we are looking for is equal to (I**3**/I0)*100 which is equivalent to the coefficient in front of I0 multiplied by 100. __** 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? Step 1. Because we are given that the image has the same orientation as the fly, one can conclude that the image is virtual. This means that the image must be on the same side as the fly. So, one can state the i-p=20 so i=20+p.

Step 2. Because the image is virtual and the magnification is positive, i must be negative. knowing that, plug the first equation into the magnification equation. So=> 2=(20+p) / p and solve for p. (p=20) This solve for the fly's distance from the lens.

Step 3. Once you've solved for p, plug it back into the magnification equation and solve for i. So=> 2= -i / 20

Step 4. Once you've solved for i, plug both p and i into the equation for the focal length. 1 / f = 1 / i + 1 / p and solve for f.

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? You would do exactly the same process as part a. Because the magnification is still positive and the orientation of the image is the same as the object, the image is still virtual, which means i is still negative, except because the image is smaller, the lens is diverging. This means that p - i = 20. Additionally, the focal length should be negative because the lens is diverging.

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. A ray 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?

Step 1. To solve this equation one must use Snell's law which states that n1 * sin(theta **I**) = n2 * sin(theta **R**).

Step 2. One must first find the angle of refraction using the equation Tan(theta **R**)= L / D and solve for the angle of refraction.

Step 3. It is given that the angle of incidence is 90 degrees, so just plug the rest in to Snell's law to find n2. __**<span style="color: #000080; font-family: 'Times New Roman',Times,serif; font-size: 14px;">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.

Step 1- First I would convert the area of the plate into meters squared. Assuming the plate is square, I would then find the length of one side of the plate (square root the area). Step 2- Then I would use the equation Change in L= L * Coeff. Lin. Exp. * Change in Temp. This would give the change in the length of one side due to the temperature increase. Step 3- Then, I would add the side length change to the original length of the side and find the new area of the plate. Step 4- Finally, I would subtract the original area from the new area to find 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.

This problem can easily be solved by using the equation P1*1V=n*R*T1. For the first part of the problem, P, V, R (constant) and T are known. This allows you to solve for n. So... solve for n and move onto the second part of the problem. Now that you have n, P2*V2=n*R*T2. The only thing missing in this equation is V2 (what you are solving for). Because the gas has not changed, n stays the same. Therefore, you can use it for the second part of this problem to solve for V2.