Tiannelli

Wikispace answers

__**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.

__Solution:__

1) Solve for Fb(max) (maximum buoyant force) with Fb(max)=Vpg 2) Solve for the weight of the logs using W=Vpg 3) To solve for the amount of children that can be allowed on the raft use the equation Fb(max)=W(children)x+W(logs), x being the amount of children allowed.
 * Use the density of the salt water.
 * Use the density of the logs.


 * 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?

__Solution:__

1) Solve for the weight of the cube using W=Vpg 2) To solve for the height that it will be covered by water use the formula W=pghA, since V=Ah. 3) Solve for h (answer in meters).
 * Use the density of the cube.
 * Use the density of the liquid.


 * //Professor's Note: Good job and a very good start !// **

__**Wiki Week 3**__ **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?

__Solution:__ there are a couple different ways to get the same answer the first way, 1) Start by converting the compressed amount into meters. 2) Next use equation w=sqrt(k/m) to solve for w. 3) Then you can use the found w to solve for v using equation v=Xm*w.

the second way, 1) Start once again by converting the compressed amount into meters. 2) Next use the conservation of energy equation by setting P.E.=.5*k*x^2 and K.E.=.5*m*v^2 equal to each other. 3) Plug in all the givens and 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. __Solution: __ 1) Start by converting the amplitude into meters and velocity to m/s. 2) Next use the conservation of energy equation by setting P.E.=.5*k*x^2 and K.E.=.5*m*v^2 equal to each other. 3) Divide the amplitude by 2 to get x and plug that and the other given amounts into the equation and solve for m.

ii. Find the period of the motion. __Solution:__ 1) To find the period(T) first find w using equation w=sqrt(k/m) 2) Next, use w to find frequency(f) in equation w=2*pi*f, reworking the equation to be f=w / 2*pi. 3) Finding f will allow you to use the simple equation T=1/f to find the period (in seconds).

iii.Calculate the maximum acceleration of the block. __Solution:__ 1) Knowing w and the amplitude, you are able to use the equation a(max)=w^2*x(max) to find the maximum acceleration. 2) Plug in w and the amplitude for a(max) in m/s/s.

**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. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">__Solution:__ <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">1) Start by converting the amplitude into meters. 2) <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">Next, find w using equation w=sqrt(k/m). <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">3) Use w to find velocity in the equation v=x(max)*w. 4) Plug in the amplitude and w to solve for volume.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">ii. Find the speed of the block when it is 4 cm from the equilibrium position. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">__Solution:__ <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">1) Convert the distance into meters. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">2) Use the equation x(t)=Xm*cos(w*t) to solve for t. 3) Plug t into the equation v(t)=-x(max)*w*sin(w*t) to solve for velocity.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">iii. Find its acceleration at 4 cm from the equilibrium position. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">__Solution:__ <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">1) Use equation a(t)=(-w)^2*x(t) 2) Plug in the w and t found in the other problem and the distance and solve for a.


 * __Wiki Week 4__**


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

__Solution:__ 1) Convert the distance to meters. 2) Set up the equation distance/phase=length of wave/2pi. 3) Solve for the length of the wave.
 * i.** What is the wave length of the wave? (Hint : one full wavelength is 2π apart !)

__Solution:__ 1) Take the given frequency and use it to solve for period using equation T=1/f. 2) Plug in the equation, phase difference/displacement time=2pi/f. 3) Solve for the phase difference.
 * ii.** At a given point, what is the phase difference (as a multiple of π ) between two ‘displacements for times 5 ms apart?

__Solution:__ 1) We know that v(t,x)=4.5 so we have to find an equation for v(t,x). 2) This is hard because the derivative cannot equal 4.5, so I'm unsure of what to do. 3) If the derivative could equal 4.5 you would take the x value given and plug into the original equation to get t.
 * 2.** At t = 0, a transverse wave pulse in a wire is described by the function y(x, t), (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: 'Times New Roman',Times,serif; font-size: 17px;">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? __Solution:__ 1) Covert u to kg/m 2) v=sqrt(F /mass per unit length) 3) Change the equation to include the Lengths. 4) The final equation is m = (v^2*u*sqrt(7/4)) / g to solve for mass.

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; 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. <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">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: 17px; text-align: left;">__Solution;__ <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">1) Use equation T=2*pi*(L/g)^.5 <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">2) Rework the problem to solve for the unknown L. (Should look like L=(T^2*g)/(4*pi^2)) <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">3) To find u the equation is simply u=m/L <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">4) Since the pendulum is non-moving, the definition of tension is F=Mg <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">5) To solve for the velocity plug in all your found data into v=(F/u)^.5

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. __Solution:__ 1) Convert the wavelengths from nm to meters. (a nanometer is 10^-9) 2) Plug the values into two equations of f=v/lambda with the different wavelengths. 3) After you find the frequency of the observer and source, plug into equation fo=fs(1/(1+(Vs/V)) 4) 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. __Solution:__ 1) Knowing B=10log(I/Io), you can plug in B and Io which is 1*10^-12. 2) Solve for I and and then multiply your finding by the number of men, 4. 3) Plug your answer back into the equation for I and then solve for B.

Professor's Note : good one on the first, and the second one refer my page !

Week eight ( 18 October, 2011)

Define polarization Polarization: The act of an electric field oscillating vertically. It can be represented by showing the directions of the electric field oscillations in a head-on view of the plane of oscillation.

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)

__Solution:__ 1) Since the original light is unpolarized you must first use the equation I=Io*(1/2) 2) The other equations involve an angle so you need to use the equation I=Io*cos^2(theta). 2) Set up your equations like I1=Io*(1/2), I2=I1*cos^2(50), and I3=I2*cos^2(50). 3) Solve for I3 by multiplying the right side of the equations for I1 to I2 and I3 together. hint: (cos^2(theta) can be plugged into a calc. as cos(theta)^2) 4) The value you are looking for will be equal to I3/Io. 5) Multiply the value by 100 to find it as a percent.

__** 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? __Solution:__  1) The distance is given so we know i+p=20cm. 2) H and H' are given so we can plug that into the equation |m|=H'/H. 3) Then m=-i/p can be used. p can then be solved for and then plugged into i+p=20. 4) When you find i just plug it back into i+p=20 to solve for p and take the magnitude since p is a distance and cannot equal zero. The equation is p-i=20 now. 5) Plug i and p into the equation 1/i+1/p=1/f and solve for f. 6) The new distance is the i and p we found added together.

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 orientation as 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? __Solution:__  1) The distance is given so we know i+p=20cm.

2) H and H' are given so we can plug that into the equation |m|=H'/H.

3) Then m=-i/p can be used. p can then be solved for and then plugged into i+p=20. i will turn out as a negative number just saying that the image is on the same size as the object.

4) When you find i just plug it back into i+p=20 to solve for p. 5) Plug i and p into the equation 1/i+1/p=1/f and solve for f. 6) The new distance is the i and p we found added together.

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? __Solution:__ 1) Theta C=Theta since the refraction is along the x-axis. 2) We know n1*sin(theta C)=n2*sin(theta), where n2 is 1 and theta is 90 degrees. 3) Theta C can be solved by tan^-1(D/L). 4) Plug theta C back into the equation and solve for n1.

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

<span style="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? <span style="font-family: 'Times New Roman',Times,serif; font-size: 18px;">__Solution:__ <span style="font-family: 'Times New Roman',Times,serif; font-size: 18px;">1) Convert all givens to meters. <span style="font-family: 'Times New Roman',Times,serif; font-size: 18px;">2) The equation needed to solve this problem is d*sin(theta)=m*lambda. <span style="font-family: 'Times New Roman',Times,serif; font-size: 18px;">3) The spacing that you are looking for is part of finding theta so you need to substitute tan^-1(y/D) into the equation d*sin(theta)=m*lambda for theta, with y being the only unknown in the equation and the part you are solving for.

<span style="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="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="font-family: 'Times New Roman',Times,serif; font-size: 18px;">__Solution:__ 1) Convert all givens to meters. 2) It is known for this problem that N1=N2, where N is the number of wavelenghts. ---> condition for the central peak 3) N1=t(n-1)+r1/lambda and N2=d*sin(theta)+r1/lambda so set them equal. 4) This leaves you with sin(theta)=t(n-1)/d, and since y=Lsin(theta) the equation you use is L*((t(n-1)/d).

**Professor's Note** : Great work and please learn and understand all the steps -- also thank you for the active participation on the wiki page !!

__**<span style="color: #000080; font-family: 'Times New Roman',Times,serif; font-size: 17.6px;">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. __Solution:__ 1) No converting is necessary if the answer is given in cm^2. 2) Use the equation delta(A)/Ao=2*(alpha)*delta(T). Alpha is the average coefficient of linear expansion. 3) Solve for 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. __Solution:__ a) 1) Use the equation PV=nRT and plug in all the givens. 2) Solve for n. b) 1) Use the equation PV=nRT to plug in and solve for a new V. P and T are new but n and R stay the same.


 * 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. __Solution:__ 1) We know Cv=(3/2)R and Q=n*Cv*(Tf/Ti). 2) First plug in the universal gas law into the first equation to find Cv. 3) Use the found Cv to plug in all the known values into into the second equation. (n=3) 4) Solve for Q.

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. __Solution:__ 1) Convert atm to Pa, and L to m^3. 2) We know the equation PV=nRT. 3) Use it to solve for T. T=(PV)/(nR). 4) We know Kavg=(3/2)k*T. 5) Plug in the found T and all the known values into the equation and solve for Kavg.

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?(b) What is the change in the internal energy of the system?(c) What is the change in the entropy of the system? __Solution:__ a1) Convert g to kg, F to K. a2) Set up an equation that looks like mc*Cc*(Tf-Tic)+ml*Cl*(Tf-Til)=0. a3) Solve for Tf when Cc=386, and Cl=128. b1) Since the volume stays constant, no work is done. Therefore, deltaEint=Q. c1) If you integrate the basic equation for entropy you get delta S=m*c*ln(Tf/Ti). c2) Use the equation for each the copper block and the lead block with the Tf we found in part a. c3) Add the two together to find the entropy of the 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

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? __Solution:__ a1) We know Ec=1-(Tc/Th), so we can use it to solve for Ec. a2) The other equation we will use to find Qh/t is Ec=W/Qh. a3) If we multiply each side by 1/t we will get Ec=(W/t)*(Qh/t). a4) Sinve P=W/t, we now have the equation Ec=P/(Qh/t). Which we can use to solve for Qh/t. b1) We know the equation Ec=(Qh-Qc)/Qh. b2) Plug in the Ec and Qh from part a and solve for Qc.