KBusque


 * 1) How many children can safely float on a raft with 6 logs? **

- Find the volume: //Pi*r^2*h*n// //n= number of logs//

//pi*.125^2*1.9*6 = .56 m^3// //- Weight of children// //200(x) = 200x//

-Weight of logs

//W= density(log)*V*g//

//800*.56*9.8 = 4390 N//

- Buoyant Force //F(B)= density(sea water)*g*v// //1024*9.8*.56 = 5619.71//

-Set buoyant force equal to weight of logs + children //5619.71= 4390+200X//
 * Solve for X **


 * 2. How far below is the bottom face of the cube **

-Determine if it floats in water a.Find the volume //(.015)*(.015)*(.015)= 3.3X10^-6//

b. Find mass //m= density of cube* volume// //(550)*(3.3x10^-6)= 1.8x10^-3//

c. Find weight //W=mg// //1.8x10^-3*9.8= 1.76x10^-2//

d. find buoyant force //F(b)= density of liquid*g*v// //1240*98*3.3x10^-6 = 4.0x10^-2//

e. set buoyant force equal to the weight of cube

//4x10^-2h=1.76x10^-2//


 * solve for h **


 * //Professor's Note : A great way to start the semester-- Good Job ! note: avoid numbers as much as you can //**
 * // //**


 * //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? ====


 * Find the energy before hitting the spring (which is maximum KE) **


 * 1/2mv^2 **


 * Then the energy of the spring compressed ( ** which is maximum PE)


 * 1/2kx^2 **


 * Then set them equal to each other and solve for m **


 * 1/2mv^2=1/2kx^2 **

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.

** Use conservation of energy **
 * E=12k(A/2)^2 + 1/2mv^2 **
 *  solve the equation so it equals m **


 * m= 3E/2v^2 **

<span style="color: #000080; font-family: 'Courier New',Courier,monospace; font-size: 14px;">ii. Find the period of the motion.


 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">Use the period formula and plugging in the spring constant and mass found in part 1 **


 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">T= 2pi(sqrt m/k) **

<span style="color: #000080; font-family: 'Courier New',Courier,monospace; font-size: 14px;">iii.Calculate the maximum acceleration of the block.

<span style="font-family: 'Courier New',Courier,monospace;">** a=K/m * A **
 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">F=ma = KA **

<span style="color: #000080; font-family: 'Courier New',Courier,monospace; font-size: 14px;">**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: #000080; font-family: 'Courier New',Courier,monospace; font-size: 14px;">i. Find the maximum speed of the block.


 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">Find the angular frequency w = sqrt(k/m) **


 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">then using SHM find x **
 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">x = Acos(wt) **


 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">then using that equation you can then find the velocity **
 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">v = -Awsin(wt) **

<span style="color: #000080; font-family: 'Courier New',Courier,monospace; font-size: 14px;">ii. Find the speed of the block when it is 4 cm from the equilibrium position. <span style="color: #000080; font-family: 'Courier New',Courier,monospace;">find acceleration **<span style="font-family: 'Courier New',Courier,monospace;">a=-Aw^2cos(wt) **

**<span style="font-family: 'Courier New',Courier,monospace;">using that plug .04 in for t and find the speed of the block ** **<span style="font-family: 'Courier New',Courier,monospace;">arccos^-1(t/v)w (arccos = cos^-1) **

<span style="color: #000080; font-family: 'Courier New',Courier,monospace; font-size: 14px;">iii. Find its acceleration at 4 cm from the equilibrium position.


 * <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">plug in .04 for t and solve for a **

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">**a=-Acos(wt) - check out this equation, you have it right in the previous part (professor) **

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">** Kassie : A wonderful effort in explaining the steps and Good Job ! **

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">** --- ** <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">** I will re-visit the page for this week's work ! **

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace; font-size: 14px;">** 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?

In order to find the mass of the object you will first need to use the following equation

V= sqrt(F/ μ)

Since the velocity and mu are given, solve for F (force)

F= v^2 *μ

Once the force has been found you can then use the following equation, and solve for m

F= m*g --> m = f/g

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="color: #0000cc; 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 <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">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;">First sense the period and gravity are given use the following equation to solve for L <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">T= 2 π sqrt (L/g)--> L = (T/2 π )^2* g <span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">Once L is found you can then find the μ by using the following equation

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;"> μ = m/L

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">Then since the mass of the ball is given you can find force using this equation:

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">F=mg

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;">And to find the speed of the transverse wave plug in F and μ to the following equation:

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 17px; text-align: left;"> V= sqrt (F/μ)

__**Week 7**__

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.

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">First you will need to find the frequency of both the observer and the source. This can be done by using this equation: <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">Frequency= C/wavelength

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">C= speed of light <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">wavelength= Plug in the wavelength observed to find the frequency of the observer and the wavelength known for the frequency of the source

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">Once you know these two frequencies you then plug them into this equation

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">F(observer) = sqrt(1-(V/c)/1+(V/c)) * F(source)

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">and solve for V (velocity) in order to find the speed of the galaxy

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.

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">In order to solve this problem you will need to use this equation:

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">dB=10log(I/Io)

<span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">plug in the decibels given for dB <span style="color: #ff00ff; font-family: 'Courier New',Courier,monospace;">since there are 4 men shouting you plug in 4*the number of decibels given for I


 * <span style="color: #000000; font-family: 'Courier New',Courier,monospace;">Professor's Note : Good job on the first one, please refer to my page for the second ! **

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?

To find the object distance the equation is |M|=H(object)/H Then plug M into the equation m=-i/p solve the equation for i Once you have found i set the distance from the image to the fly equal to i+p Plug in the i you found in the equation above, and solve for p plug in your value for p and solve for i Now that you have your i and p values plug them into the equation 1/i + 1/p = 1/f that is your focal length

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?

Find your new object distance by using the same equation |m| = H(object)/H

To find your new focal length: 1) plug i into the equation i+p(found in part a) = distance from fly to 2nd lens 2) plug i and p into the equation 1/p + 1/i = 1/f to find your focal length
 * Professor's Note: Please visit my page and get a better understanding on the virtual images and associated lenses. **

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?

First find the tan of angle 1= L/D Take the index of refraction of air to be n2 The refractive angle is 90 degrees plug these values into this equation n1=n2(sin(L/D)/90) Solve for n1

__ **<span style="color: #000080; font-family: 'Times New Roman',Times,serif; font-size: 14px;">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.

∆A = β A ∆T

Plug the original area in for A β is the coefficient of linear expansion T is how much the temperature increased by then the change in Area can be found