sbains

__**Quiz 4**__ 1 a) You are given, the distance between the points, the out of phase value, and it is known that one full peroid is equal to 2pi. It is asking for the wavlength lambda. Using this information a ratio must be set up by 2pi /point distance=out of phase value/lambda. Solve for wavlength lambda

b) For this part you are given time between two points, you can calculate the period using the 1/f equation, and lambda from part 1. It is asking for the phase difference. Using this information you can set up another ratio: (time between two points)/(phase difference)=(period calculated)/(phase difference). Using that you can now solve for the phase difference.

2) It is asking for the positive x direction for a given velocity. By taking the derivative of the y(x,t=0) with equation it is now for velocity instead of displacement. This is done because the time is wanted when given velocity. Setting the derived equation equal to the given and solve for x. Now having the x value for the velocity, plugging that x value into the original displacement equation gives a final value for x. __**Quiz 5**__ 1) The very first thing to do is for this problem is to draw a diagram. You are given the length of the string in equal parts as 3L/4 and the sum of one force(tension) is T*sin(theta), and since there are two strings it becomes, 2T*sin(theta). The force tension can be written as mg. You are given the length of the string, the fact that they are divided equally as L/2 on each side, the mass per unit length is given, and the wave speed is given. After all the values are converted to thier si units. The angle can be calculated by using trigonometry. Now using the equation sin(theta)=sqrt(7L/8)/(3L/8)) and then using the equation for V being V=sqrt(F/mass per unit length), the force is the quantity of 2*mg because of two strings, multiplied by the equation calculated above for sin(theta). Using that equation it is now possible to sovle for m.

2) For this pendulum problem, the best equation to use is T= 2*pi*(L/g)^0.5, Since you are given T, and g is a known quantity you can now solve for L from that equation. Using L, we can now solve for u, using the equation u=m/L. The mass (m) and L are known. The pendulum is stationary, when this is the case the equation to be used is F=mg because there is only a dead wieght of the ball and gravity that matters for the force. Now since you have F and u, the equation v=(f/u)^0.5 can now be used to solve for velocity (v). __**Quiz 6**__ 1) Convert given wavelength to m instead of nm 1nm=1*10^-9m using these values for wavelength the frequencies can be determined using f=vlight/wavelength vlight=speed of light 2.99792*10^8 the frequency should be gotten for the galaxy and the laboratory now using the equation: f ' = f*(1/(1-vlight/vs) f'= is the frequency observed from the laboratory f= is the frequency from the galaxy solve for vs the velocity of the galaxy in the line of sight relative to the earth

2) using the equation: B=10log(I/Io) and using (I4 = 4 I1) because now it is 4 times the intensity of one man B=10(log(I/Io)+log(4)) 10*log(I/Io)+10log(4) since it is given 20 db is the loudness then, 10*log(I/Io)=20 so 20+10*log (4) and 10*log(4) can be calculated Polarization- when rays of light or electromagnetic radiation are restricted to directions of vibration 1) Equation I=.5 Io illustrates the unpolarized light before entering the polarizing sheets. When the light does enter through the sheet the unpolarized light becomes polarized allowing us to use I=Io*cos(theta)^2 So for I1= Io*cos(theta1)^2 for I2= I1*cos(theta2)^2 I3= I2*cos(theta)^2 Using the I3 equation (since this is the last sheet) and plugging in for I2 which equals I1*cos(theta2)^2, and then within that plugging in for I1 which equals Io*cos(theta1)^2. You end up with I3=Io*((cos(theta1)^2)*(cos(theta2)^2)*(cos(theta3)^2) I3 =(some value)*Io the percentage of initial intensity is I3/Io=(some value)(100%)
 * Professor's Note : good job on both ... keep it up!! **
 * __ Quiz 7: __**

1a) Variables: di=image distance do=object distance m= lateral magification hi=height of image ho=hieght of object f=focal length 1) The image produced is a virtual image because the image and the object have the same orientation, so the image is on the same side of the lens as the object and m is positive
 * __Quiz 8__**

2) Since it is stated that the image is 2 times the size of the object the magnification, so the type of lens is a convex lens or converging lens. Since m=-hi/ho. M becomes -2hi/ho. So m=2. This means that 2=-hi/ho which also is equivalent to 2=-di/do and using the absolute value for di, becomes 2=di/do.

3) It is stated that the distance between the image and the fly is 20 cm. So since the image is bigger than the object so the second equation is di-do=20

4) So solving for di in the second equation, becomes di=20+do. Plugging that in for di in the first equation will get a value for di. Using that value plugging into either equation will get you a value for do.

5) using the equation for a thin lens 1/f=1/di+1/do makes it possible to solve for f now that you have di and do.

b) 1) This time the fly goes sits in front of a second lens with the image and the object in the same orientation as before and the distance between the fly and the image is 20 cm as before, but this time the image is 1/2 the size of the object so m=0.5hi/ho, the image is virtual so m is positive and the lens type is concave or diverging.

2) m becomes 0.5 and the equation is 0.5=di/do using the absolute value of i again.

3) Because of the magnification being 0.5, the image is smaller than the object. This means second equation becomes do-di=20 cm and the image is on the same side of the lens because it is virtual but it is in front of the object because the image is smaller.

4) Using the two equations you can get values for both di and do, and from there using the 1/f=1/di+1/do equation to get the focal length

Professor's Note : good explanation (please understand the steps !)

2) For this problem it is asking for the index of refraction of the liquid. It is best to use the Snell's Law equation which is n1*sin(theta1)=n2*sin(theta2)

n1 is the medium initially and it is 1 because the medium is air initially and index of refraction of air is 1 theta 1 is referred to as the incidence angle, the angle the light ray has initially it is 90 in this case because the light ray is perpendicular to the surface of the tank

n2 is the index of refraction of water and is the value we are trying to get

theta 2 is the refraction angle, the angle the ray produces after reached the other medium (the bottom right angle where the ray if light ends) it is not given but it can be calculated by using the lengths given and using trigonometry to get it it becomes tan(theta2)=D/L. Solving for theta2 will get a value for theta2.

Now all that is needed to do is plug in all the values solved for/given and solve for n2 using the snell's law equation

1) a. First the formula d*sin(theta)=m*lamda is used because the spacing between bright fringes m=1 and m=2 is wanted. Solve for theta 1 by using m=1 for m, and the slit seperation for d, lamda is given. Solve for theta 2 by using m as 2 and rest as the same as for theta 1. Using the equation tan(theta)=o/a=y1 or y2/a. Using each of the angles found above, y as the distance b between fringes and a as the distance between the slit and screen which is given. The distance between fringes (y) can be found. y1 is using the first angle (theta 1) and y2 is for the second angle (theta 2). As a result y2-y1 is the distance.
 * Quiz 10**

2) The equation N2-N1=(L/lamda)(n2-n1) is used in this situation. The value wanted is N2-N1 which is the distance of wave number. N2 is the index of refraction for the plastic and n1 is the index of refraction for air. And to find the vertical distance y that the interface pattern moved use the equation (N2-N1)*(distance between fringes).

Professor's Note : First part is correct and refer to KarenB's page for the complete answer for part 2 or HW solutions.

1) This problem deals with thermal expansion in terms of area. It's giving you a inital area, and it also gives you a change in temperature that the plate is subjected to. It wants to know what the increase in area or change in area that is caused by the change in temperature. The equation (delta V)=V*beta*(delta T) is used because this explains the relationship between the change in volumedue to thermal expansion. Even though this problem is dealing with area and not volume this equation is used instead of the linear expansion equation because in this problem multiple dimensions are moving as opposed to one dimesion in linear expansion.
 * Quiz 11 **

However, when dealing with volumes, one relationship to know is that beta=3*alpha, but when dealing with area the relationship is beta =2*alpha because areas usually only deal with 2 dimensions as opposed to three with volumes. Using the this relationship we can now plug in the relationship to the formula and just change delta V to delta A for area. Looking like: deltaA=A*(2*alpha)*(deltaT). Change in temperature is given (deltaT), alpha is a constant which is also given, and the original area is given A. Plugging those values into the equation will give you change in area (delta A) which the problem is asking.

2a) This problem, calls for the formula for ideal gas, pv=nRT because it is giving you all the information for the formula and it is asking for the moles which is known as variable n. It's telling you that the ideak gas a has a certain temperature (T) and that it has a pressure (p), in kPa which needs to be converted to pa (si units) and it is giving you the volume (v). You also know the variable R which is the gas constant as 8.31 J/mol*K. Plugging all of these variables in and you are left with n which is the amount of moles. b. For the second part you are given a new pressure value (also needs to be converted to Pa) and a new temperature and you are asked to find the volume. The thing to know for this problem is that the only two factors that change a gases' volume is the temperature and the pressure. All the other variables like number of moles (n) and the gas constant (R) stay the same for the gas. So plugging in n and R from the first part of the problem and the new values for temperature (T) and pressure (Pa). You are left with v, solve for v and that is the volume.