RajinR

Quiz Questions

__**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.
 * 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 1000 kg /m3, that of liquid is 1300 kg/m3. How far below the liquid surface is the bottom face of the cube?

1). Step :1 Knowns: Density = 800 kg/m^3, sea water is 1024 kg/m^3, 6 logs, F= 200N, log diameter = 0.25 m, log length = 1.90 m - The total volume of the raft needs to be determined. volume = 6 x (((0.125^2 x pi)) x 1.9) = 0.55967m^3. - Then the mass of the raft must be determined. Mass of raft = (.55967 x 800), = 447.735kg Step:2 0.55967m^3 of water can be displaced by the raft, when just fully submerged. (.55967 x 1024) = max. 573.1kg. can be supported. (573.1 - 447.735) = 125.367kg. max children mass, x 9.8 = 1,228.597N. (1,228.579/200) = 6 children max.
 * Equation of a Cylinder - V= pi*r^2*h

//Professor's Note: work is correct, please avoid using numbers in the future. Just give the steps to solve the problem. A very good start !//

**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?


 * Given: m= 1000 kg **
 * k= 5 x 10^6 N/m **
 * X(m) = .0316 m **

1. The energy of compression is (1/2)kx^2. You've been given k and x. Calculate it. That came from kinetic energy (1/2)mv^2. Set this expression equal to the energy, plug in m, and solve for v.

i. Calculate the mass of the block. ii. Find the period of the motion. iii.Calculate the maximum acceleration 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.

i). Energy of compression is (1/2)kx^2. Calculate it at the maximum compression (10 cm) and halfway compressed (5 cm). The difference is kinetic energy. Set that equal to (1/2)mv^2 and solve for m.
 * Given: k = 6.5 N/m**
 * X(m)= .1 m**
 * mass = ?**

ii). Period: Plug the known values into the formula for period. (2pi/T)

**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 !)

Step 1: Write down all of the given values from the problem. Step 2: Plug in and use the equation Lambda = 2 pi*(x/phi) to solve for the wavelength (lambda).

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: Find the period of the wave by using the equation: T = 1/f Step 2: Next, divide the period by the dispalcement. Step 3: Then, take the answer form the previous step and divide it by 2 pi.


 * Professor's Note: ** You have done a great job here and I am proud of you. Explanations are great and now you are thinking before touching the calculator,,, Bravo !!

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) )

**Fifth week (26th September, 2011)**

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: First, draw a free body diagram of the mass hanging from the string. The length of the string is given as 3L/4. Step 2: From the free body diagram, the equation: 2* T * sin(theta) can be obtained. Step 3: Use the equation and the given length of the string to compute the angle that the string forms. Step 4: Next, take the value from the previous equation and plug it into the equation: v = sqrt(T/mew) to solve for the 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.

Professor's Note : good one on the first, and the second one incomplete !

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.

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.

The total level in dB is the level of one sound source plus the increase of level in dB. One man in the distance has the loudness level of 20 decibels. 4 men have the increased total loudness of Δ L = 10×log (4) = 6 dB more + 20 dB

Professor's Note : Please refer my page !

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)

Polarized light waves are light waves in which the vibrations occur in a single plane. The process of transforming unpolarized light into polarized light is known as polarization.

we use the equation for half the intensity because we are trying to solve for unpolarized light that is polarized. I = .5 io

__ **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?

First Step: Identify knowns from the problem.
lamda = 643 x 10^-9 m d = .15 x 10^-3m L = 1.45m m = 1 Theta = ? Y = ? Second Step: Use the equation d*sin(theta)=m*lamda, and can solve for theta. Third Step: Use the equation sin(theta)=Y/L and solve for Y, which is the distance in between the bright fringes.

Professor's Note : Good job on the first part. What about the second part ?

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.

__** 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 Analyze the problem and look at your knowns and determine your unknowns. Step:2 Since, the problem provides the linear expansion, one can conclude that the equation: Delta(l) = L (alpha)(delta t), will be used. Step:3 Rearrange the equation to suit the problem. (delta A) = A(2xAlpha)(delta t) Step:4 Solve for the change in area, Delta A.

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.

Step:1 Analyze the problem and look at your knowns and determine your unknowns. Step:2 Since, the problem asks for the number of moles you can use the equation pv=nRT to solve for n. Step:3 I'm not quite sure how to do this.

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.

Step:1 Analyze the problem and look at your knowns and determine your unknowns. Step:2 Since the problem states that the gas is at constant pressure,you can use the constant pressure equation to solve for the heat transferred to the gas. Q = nCpDeltat(T) Step:3 All values are in proper SI units so no conversions needed. Just plug the values into the equation and solve for the heat transferred.

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.

Step:1 Analyze the problem and look at your knowns and determine your unknowns. Step:2 Since the problem asks for the average translational kinetic energy of a single molecule, we know that the equation: Kavg = (3/2)kT will be needed. Step:3 Note that the temperature change is not given in the problem so the equation: PV = N k T can be used to find the temperature of the vessel. Step:4 When the temperature is found, simply substitute the values into the equation: Kavg = (3/2)kT to find the average translational kinetic energy.

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?

Step:1 Analyze the problem and look at your knowns and determine your unknowns. Step:2 P = 600 W Hot Reservior TH = 100 degree celsius = 373 k Cold Reservior TL= 60 degree celsius = 333 k Step:3 Find the efficiency of the engine. Step:4 Then find Abs(QH) Energy taken in as heat Step:5 Then Find Abs(QL) Heat energy exhausted