ABroatch




 * Professo's Note : Great job -- everything is pretty clear and love the diagram. Wonderful !! Thank you for actively participating in the Wiki effort !! **
 * Note: N1 = N2 is the condition that is satisfied by the central peak __not any maximum__ **



Comment on 1st Answer : When the image has the same orientation, it forms in the same side as the object.



professor's Note : Great job ! good answers -- thank you fo doing such a wonderful job !

Week #6: No Wiki Problem.





**Professor's Note:** Liked the new style !! Good job !! My only concern is (don't like to see the numbers so much)... Did I mention, that I liked the style?

__Problem#1:__ Write down the given information(mass, spring constant, Xm) Draw a picture of the situation. I drew a picture of a block attached to a wall 3.16cm away with a spring. In this case, the bumper acts as the spring.
 * WEEK THREE: **

In order to find the speed of the car before the impact, I used the equation Vmax= Xm*w and w=sqrt(k/m) to find that Vmax = Xm*(sqrt(k/m)) *Note w=omega

You could also solve for V using the equation .5kx^2 = .5mv^2 Since there is no energy lost during the collision

__ Problem#2: __ Write down the given information (spring constant, Xm, Vmax) Draw a picture of a block attached to a wall 10 cm away with a spring. Label the equilibrium point.

i. To calculate the mass of the block I calculated the Potential energy of the block at 5 cm and 10cm then subtracted the values to obtain the resulting Potential Energy. [ .5k(.1)^2]-[.5k(.05)^2]=PE

Next, use the equation PE = KE to solve for m PE=.5mv^2 remember to change the velocity to meters/second.

ii. To find the period of the motion, we can use the equation T=2*pi*sqrt(m/k)

iii. Use the equation Amax = Xm*w^2 and w= sqrt(k/m) to find the maximum acceleration Thus: Amax = Xm*(sqrt(k/m))^2

__ Problem#3 __ Write down the given information (mass of block, spring constant, Xm) Draw a picture of a block attached to a wall 8cm away with a spring.

i. To find the maximum speed of the block, use the equation Vmax= Xmw = Xm(sqrt(k/m)) ii. The speed of the block will be its greatest 4cm away from its equilibrium position. So V=Vmax, which we just calculated iii. The blocks acceleration will be its greatest 4cm away from its equilibrium position. To find Amax, use the equation Amax = Xm*w^2 = Xm(sqrt(k/m))^2

__Problem #1__
 * __WEEK TWO:__ **

To start, write down the given information ( Number of logs, Dimensions of log, Density of logs and sea water, and average weight of each child) Since the raft needs to float, we can conclude that the bouyant force and the total weight need to be in equilibrium. Thus: Fb = Wraft + Wkids

Now need to determine Wraft (weight of raft), Wkids ( weight of kids), Fb (Bouyant force)

To Find Wraft: Wraft = v*p*g multiply the volume of raft (6*pi*r^2*l), density of wood, and gravitational force **Remember to use the radius in this equation, not the given diameter**


 * To Find Wkids: **
 * Wkids = Waverage * n **
 * multiply average weight of each child by n, since we need to determine how many children the raft will allow **


 * To Find Fb: **
 * Fb = v*p*g **
 * multiply volume of raft (6*pi*r^2*l), density of liquid (salt water), and gravitational force **


 * Now we can plug these values into the equation Fb = Wraft+Wkids and solve for n. **
 * (I got n=6) **


 * __Problem #2__ **
 * Write down given information (density of wood, density of fluid, length of cube) **
 * Draw a diagram of the cube floating in the water and locate the distance h we need to calculate **


 * Since the cube is floating, the Bouyant force and the Weight of the cube are in equilibrium. Therefore: **
 * Fb = Wcube **


 * you can re-write this equation as **
 * pl*g*h*A = v*pc*g **


 * where **
 * pl = density of liquid **
 * g = gravitational force **
 * h = distance from water surface and bottom of cube **
 * A = (length of cube)^2 **
 * v = (length of cube)^3 **
 * pc = density of cube **

Professor's Note : Good job ! very nicely done and a great way to start the semester -- KIU**
 * now we can solve for h using the given information and the above equation. **
 * (I got h=0.0067 m ) **