Mcordone

__** Week 12 **__

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.

Given :

change in temperature = 100K Cp=28.8 J/mol*K R= 8.31451 J/mol*K n= 3 mol

1. since its at a constant pressure, then the equation needed is Q=(n)*(Cp)*(change in temperature). then Q can be solved for.

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.

Given:

V= 5L n= .125 Mol P= 1.5 atm K= 1.38066 x 10^-23 J/K R= 8.31 J/K

1. use the equation PV=nRT to solve for the change in temperature. 2. then use the equation Kavg= (3/2)*K*T

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? What is the change in the internal energy of the system? What is the change in the entropy of the system?

Given:

Temp initial of copper = 150 F temp initial of lead = 100 F mass of copper = 50 g mass of lead = 100 g

1. convert everything to SI units 2. the equation (mc)*(Cc)*(change in temperature of copper)+(ml)(*Cl)*(change in temperature of lead)=0, and solve for the final temperature. 3. Q=change in Eint, because the pressure is constant.

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?

__** 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.

Given:

area of plate = 95 cm^3 coefficient of expansion = 5 × 10 − 6 (ºC)-¹ change in temperature = 112 C

1. to find the length of the plate, take the square root of 95. 2. once L is found, you can use the equation; (change in L)=(L)*(coefficient of expansion)*(112 C) 3.the change in length can then be added to the original length, because that is how much the plate expanded, and the new area can be found by squaring the new length. 4. once the new area is found, minus that from the original area, and that is your 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.

given:

P= 100000 Pa V= 2.5 m^3 n= ? R= 8.31 J/molK T=283.15 K

a) once all the information is plugged in then the amoun of moles can be solved for with PV=nRT, because it is an ideal gas.

given:

P= 300000 Pa V= ? m^3 n= answer from a R= 8.31 J/molK T=303.15 K

b) if the amount of moles from part a is used, and the temperature and pressure is changed, then the new volume can be solved for with the same equation.

__** Week 10 **__

1. 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=1.45 nm lambda= 643 nm d=150000 nm m=1 (because its the spacing between each bright fringe) theta=? Y=?

1.first you need to find theta, with this equation, d*sin(theta)=m*lamnda. you have everything, and can solve for theta. 2. because theta is small, the equation sin(theta)=Y/L. Y is the distance between the bright fringes. so plug in everything, and solve for Y. note Y will be in nm, because everything was converted to nm in the beginning. professor's Note : Good job and the correct steps !! 2. 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.

Given: t (thickness of plastic) = 480 nm n of plastic = 1.5 L= 1.45 x 10^9 d=150000 nm

so with this information, this equation can be used to solve for y, the distance the central max moves upward due to the plastic.

y=L*(t*(n-1))/d), remember the answer will be in nm, because everything was converted in nm to start


 * professor's Note : correct final equation but more explanation please :) **

__** Week 9 **__

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?

given:

m=2 p+i= 20 cm (not correct !) f=? p=?

1. so you can get 2 equations from the given info in the solve function in the calc, and solve for i and p. m=-i/p and i+p=distance between object and image. it will look like this. solve(-2=-i/p and i+p=20, {i,p}). 2. once you have a value for i and p, you can plug them into this equation to solve for the focal point. 1/i+1/p=1/f. 2. the object distance is just the p value that was solved for before.

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?

given:

m=.5 p+i=20 cm (not correct !) f= ? p= ?

1. so you can set 2 equations in the solve function in the calc, and solve for i and p. it will look like this. solve(-.5=-i/p and i+p=20, {i,p}). 2. once you have a value for i and p, you can plug them into this equation to solve for the focal point. 1/i+1/p=1/f. 2. the object distance is just the p value that was solved for before. **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.

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?

1. to find the angle on incidence, you can solve for theta because you have the two side lengths of the right triangle, with arctan(1.1/.85) 2. the other angle is just going to be 90 degrees because its along the horizontal axis. 3. then with snell's law, you can solve for the index refraction of the water. n1*sin(theta 1)=n2*sin(theta 2). solve for n2, n1 will equal one, because the index of air is one.

__** Week 8 **__

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)

1. so all thetas are given to be 50 degrees. the equation for the first filter will be I1=.5Io, because the light coming in is unpolarized. for the next filter it will be I2=(I1)(cos^2(theta 2 - theta 1)), and the third filter will be I3=(I2)(cos^2(theta3 - theta 2)). this is because the light coming in is already polarized into the second two filters. then you multiply it by 100 to get a percent.

__** Week 7 **__

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.

Given: lambda source= 3.43X10^-7 m Lambda observed = 4.62X10^7 m speed of light 2.99792X10^8 m/s speed of sound = 340 m/s

1. so you have the velocity and lambda of the two waves, so with that info, you can solve for the frequencies with f=v/lambda 2. with the two frequencies, the equation f(observed)=f(source)[1/(1+(Vsource/Vsound)] can be used. you have everything except the velocity of the source, which is what you are solving for.

2) 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.

Given: loudness of one man: 20 dB //I//o(intensity knot): 1X10^-12 w/m^2 //I//(intensity):? loudness of four men:?

1. use the equation loudness=10*log(//I/I//o) to solve for //I//. 2. once //I// is found, the plug that into the equation multiplied by four, because now there are four men, and solve for the loudness. 3. so the equation will like like this; loudness=10*log(4*//I/////Io),// then just solve for loudness.

Professor's Note : Very good answers !

__** 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?

1**.** Given:

u= .008 kg/m v= 60 m/s m= ? g= 9.8 m/s^2 distance from wall to wall = 3/4L distance of string from mass to wall = 1/2L tension = w/sin(theta) theta= ?

2. first you need to find theta so you can calculate the vertical component of the tension with t=w/sin(theta). so if you draw a diagram, you can solve for theta. if you draw a line in the middle of the two walls going downward, the the hypotenuse will be L/2, and the adjacent side will be .5(3/4L). with this, the arccos can be used to solve for theta.

3. once theta is found, then the tension can be solved with m still in, and tension = m(g)/(sin(theta)) can be plugged into v=sqrt(tension/u). then everything except m is given, and you can solve for 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.

1. given:

mass= 5 kg T= 2 seconds g= 9.8 m/s^2 length= ? v= ?

2. so to find the length of the string, use the equation T=2pi*sqrt(L/g). everything is given except length, which is what you are solving for. you will need the length later to solve for u, which is mass/length.

3. once you have the length you can solve for u, then use v=sqrt(tension/u) to solve for velocity.

__** 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 i. What is the wave length of the wave? (Hint : one full wavelength is 2π apart !)

**i.** 1. Given distance between points= .05 m out of phase=pi/3 one full period is 2pi

2. so with this information, you can set up a proportion and solve for lambda. so after cross multiplying you get (2pi)(.05)=(pi/3)lambda. then solve for lambda.

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

1. Given lambda=solved in part one period=1/frequency time between two points= .005

2. so you can set up the proportion (time between the two points)/(change in phase)=(period)/(2pi). so when you cross multiply, you get (.005)(2pi)=(1/35)(phase difference).

2) 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) )    1. take the derivative of the position function, because that will give you the function for velocity. the velocity is given as 4.5m/s, so set that equal to the equation, and solve for x.    2. once x is found, plug that back into the position equation, and solve for y(x,t).

__** 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?

1. Given: m=1000 kg k=5000000 n/m Xm=.0316 m

1. so KE=PE because no energy is lost in the impact.

2. so the equation 1/2(m)(v)^2=1/2(k)(Xm)^2 can be used. now just 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.

**i. ** Given: m=? k=6.5 n/m Xm=.1 m v=.30 m/s <span style="color: #000000; font-family: 'Times New Roman',Times,serif; font-size: 110%;">w=? <span style="color: #000000; font-family: 'Times New Roman',Times,serif; font-size: 110%;">x=.05 m

<span style="color: #000000; font-family: 'Times New Roman',Times,serif; font-size: 110%;">1. so to find mass the equation w=sqrt(k/m) can be used, solving for mass.

<span style="color: #000000; font-family: 'Times New Roman',Times,serif; font-size: 110%;">2. first omega needs to be found with the equation v=(w)(x). v and x are given.

<span style="color: #000000; font-family: 'Times New Roman',Times,serif; font-size: 110%;">3. then with w found, w and k can be plugged into w=sqrt(k/m), and solve for m.

**ii.**

1. the equation for the T(period) is T=1/f. to find f(frequency) use the equation f= (w)/(2*pi)

2. w was found in part i. once f is found, it can be plugged into T=1/f

**iii.**

1. the equation for max acc. is Am=(w)^2*(Xm). Xm and w were previously found, so the just need to be multiplied.

<span style="color: #0000ff; font-family: 'Times New Roman',Times,serif; font-size: 110%;">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="color: #0000ff; font-family: 'Times New Roman',Times,serif; font-size: 110%;">i. Find the maximum speed of the block. <span style="color: #0000ff; font-family: 'Times New Roman',Times,serif; font-size: 110%;">ii. Find the speed of the block when it is 4 cm from the equilibrium position. <span style="color: #0000ff; font-family: 'Times New Roman',Times,serif; font-size: 110%;">iii. Find its acceleration at 4 cm from the equilibrium position.

**i.** Given: m= .4 kg k= 12 n/m Xm= .08 m

1. to find max speed, omega is needed. the equation for max speed is Vm=(w)*(Xm).

2. so to find w, use the equation w=sqrt(k/m). k and m are given.

3. then plug w and Xm into Vm=(w)*(Xm).

**ii.**

1. so if the block is 4 cm from equilibrium, x(t) will = .04 meters. the use the formula x(t)=(Xm)cos(w*t). w was found in part i. solve for time.

**iii.**

1. the equation for acc. is a(t)=(-w)^2*(x). w was already found in part i, and x is .04 meters.

__** Week 2 **__

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. find the volume of a single log, with the formula (pi)(radius^2)(height). 2. then multiply this answer by 6, because there are six logs. 3. use the formula (volume)(density)(gravity). the volume was found in step 2, density given for the wood, and use 9.81 for gravity. now this is the total weight for just the raft, without the children. 4. now find fB, using the same formula, but with the density of the salt water. so you have (volume of the six logs)(density of the salt water)(gravity). 5. this number is going to be bigger than the first, so subtract the smaller number from the bigger answer. this will be how much weight the raft can carry before sinking. 6. so divide this number by 200, because that is how much each child weighs, the answer a rounded down to the nearest whole number because you can't carry a decimal of a child. <span style="font-family: 'Times New Roman',Times,serif;">// Professo's Note : Very nicely done Mike - Good job ! //

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. the equation for this problem is (volume of liquid)(density liquid)(gravity) = (volume of cube)(density of cube)(gravity). 2. for the liquid side of the equation, you will find the volume by (length)(width)(height). but the height will stay as h, because that is what we are solving for. 3. the right side of the equation, all of the information is given to complete. the volume is (length)(width)(height), the density is given, and gravity is 9.81. 4.so now just solve for h, and this is your answer. the answer will be less than 1.5 cm because this is how tall the cube is.

<span style="font-family: 'Times New Roman',Times,serif;">// Professo's Note : Again, what a great way to start the semester, KEEP IT UP // <span style="font-family: 'Times New Roman',Times,serif;">// - //