karen+z

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. **

steps: 1. Convert the wavelength given into its si unit. 1 nanowavelngth=1*10^-9wavelength 2. You are given the wavelength of the observer and source. Now convert the wavelengths into frequency. Frequency and wavelength of light are related through the speed of light therefore you can use f*wavelength=speed of light, to obtain the frequency.

3. You are asked to find the speed of the galaxy in the line of sight to the earth. The source (light of the galaxy is coming towards the oberver(relative to the earth) Earth being stationary. You can use the doppler's effect:

frequency(observer)= frequency(source)*[1/1-{V(source)/V(speed of light)}] speed of light is given.

4. Now, plot the given and solve for the unknown source.

**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 wave**s.

equation to solve for decibel level: β = 10db log(I/I(knot)) Steps: 1. you are told that the four men shouting are equally powerful when shouting at a distance, so then one men shouting from the same distance leads to conclude the following. intensity of a man shouting (I(1))=4*that intensity (for four men shouting). (4I(1))

2. you are given the decibel level of a man shouting, being 20dB. you can write the following as:

20dB=10dB*log(I(1)/I(knot)),> beta(1)

3. Based on the above information and that there is no interference, you have the following:

(dB of 4 men shouting)=10dB*log(4*I(1)/Io) you can simplify the following using some logarithm rules (multiplication in this case) and have the following:

rule of multiplication ex. log(4*5)=log(4)+(log(5)) if and only if the base of the logs are the same. **Log=log(base of 10)**
 * dB of 4 men shouting= 10dB*log(I(1)/I(knot))+ 10dB*log(4)); you know beta(1)= 20dB so you can rewrite the following to:**


 * dB of4 men shouting= 20dB+ 10db*log(4)> voila... solve for the level decibel of the four men shouting**

Professor's Note : Very good 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.

State the given: **Density of the logs (p)**= 800kg/m^3 **average weight of a child**= 200N **diameter**=.25m; **radius=** .125m **number of logs=**6 **length (l)=**1.9 m

With the following you can calculate Volume, which is volume of a cylinder times the number of logs:

**V= (pi*r^2*l)*number of logs**

**We are ask to find how many children can safely afloat on the raft in sea water.**

**you can solve for the bouyant force, weight of the log:** **F(b) or bouyant force= Volume * fluid density * 9.81 meters per second squared.** **Weight of the log= volume * density of the log* 9.81 m/s^2**

**the whole system is going to be in equilibrium since we are look for the amount of kids that can be on the log, that wont cause the log raft to sink, there is no motion, so with the following information we can do the following**

**the sum of the force in the y direction, since there is no force being applied in the x direction:**

**F(sum)= - F(bouyant) + W(log) +W(of the children which is unkown)** <span style="font-family: 'Times New Roman',Times,serif; font-size: 90%;">**0=** **- F(bouyant) + W(log) +W(of the children which is unkown)** <span style="font-family: 'Times New Roman',Times,serif;">** F(bouyant) - W(log) =W(of the children which is unkown )** <span style="font-family: 'Times New Roman',Times,serif;">**[F(bouyant) - W(log)]/ W(of the child) = # of children**

<span style="color: #2a00ff; font-family: 'Times New Roman',Times,serif;">**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.**

<span style="font-family: 'Times New Roman',Times,serif;">**Given:**

<span style="font-family: 'Times New Roman',Times,serif;">**edge of the cube (l) = 0.015** from there you can derive the volume and surface area; **V= l * l * l; A= l^2** <span style="font-family: 'Times New Roman',Times,serif;">**density of the wood (p)= 550 kg/m^3** <span style="font-family: 'Times New Roman',Times,serif;">**density of the liquid (p)= 1240 kg/m^3**

<span style="font-family: 'Times New Roman',Times,serif;">**with the following we can solve for the Weight and then solve for F(bouyant) of the floating object on the liquid:** <span style="font-family: 'Times New Roman',Times,serif;">**W (wood)=mg=pgV=density of the wood*9.81 m/s^2*V**

<span style="font-family: 'Times New Roman',Times,serif;">**we know that F(bouyant):** <span style="font-family: 'Times New Roman',Times,serif;">**F(bouyant)=density of the fluid*g*h(area of the object)** <span style="font-family: 'Times New Roman',Times,serif;">**and also F(bouyant)=mg so we can write the following :**

<span style="font-family: 'Times New Roman',Times,serif;">**W(wood)=F(bouyant)** <span style="font-family: 'Times New Roman',Times,serif;">**W(wood)=** **<span style="font-family: 'Times New Roman',Times,serif;">density of the fluid*g*h(area of the object), isolate the h and solve, and that should be how far below the liquid surface is the bottom face of the cube. ** <span style="color: #800000; font-family: 'Times New Roman',Times,serif;">//Professor's Note: **Very good answer (s), a great way to start the semester, KIU**//

<span style="color: #800000; font-family: 'Times New Roman',Times,serif; font-size: 140%;">//**Third Week**//
 * 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?

Step one, state the given and unknown:

---> mass=1000 kg --->similar to a spring constant (k)= 5 x 10^6 N/m --->amplitude or X(m)= 3.16 cm convert to .0316 m --->car is brought to rest we know V(final) = 0 --->neglect energy lost during impact
 * Given:**

Unknown:
 * the initial velocity of the car before impact**

The best formula to solve the following is the mechanical energy before and after the impact, therefore:

before impact-> KE +U(t) = KE +U(t)<--after impact


 * we know that before the impact there is no resonance and that potential energy is transmitted once the car hits the brick.**


 * Following the conservation of energy, that states if the forces work within an isolated system than the system cannot change,**


 * KE=U(t)**
 * 1/2 * m * V^2 = 1/2 * k * X(m)^2**
 * the car is transferring its kinetic energy into potential energy once its impact with the wall.**


 * now plot in the given and solve for the unknown V.**

<span style="font-family: 'times new roman',times,serif; font-size: 120%;">**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.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">Before beginning to answer part i, state the given and unknown: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**spring constant (k**)= 6.5 N/m <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**X(m) aka amplitude** = 10 com convert to .1m
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">mass=? **
 * V(max)= V(measured) * 2,** because it gave us the speed halfway through and doubling the velocity will give us the maximum velocity the block will be traveling.


 * i. Calculate the mass of the block.**


 * we know that V(max) is equivalent to the amplitude times omega (angular velocity) and that the spring constant divided by the angular velocity squared will give us the mass. So it is as follows:**


 * V(max)=X(m) * omega--> omega= V(max) /X(m)**
 * therefore, omega=sqrt(k/m)-> mass= spring constant (k) / omega^2, and solve for mass **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">ii. Find the period of the motion.
 * From the previous part we solved for mass, but along that we also solved omega.**
 * Omega is also equivalent to 2pi/T(period)**

iii.Calculate the maximum acceleration of the block.
 * so, T(period)= 2pi/omega **

**a(max)=omega^2 * X(m)we know omega and the amplitude was given, just solve for a(max)**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">**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="font-family: 'Times New Roman',Times,serif; font-size: 120%;">State given and unkowns: <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">mass=.4 kg <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">spring constant = 12N/m <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">X(m)=.08m

i. Find the maximum speed of the block.

**V(max)=X(m) * omega> omega= sqrt(k/m)**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 130%;">ii. Find the speed of the block when it is 4 cm from the equilibrium position.


 * we know that x(t) = X(m) * cos(omega * t), we can solve for t, since we are given displacement and we already solved omega from the previous part.**
 * After solving for t, use the following equation**
 * V(t) = X(m) * omega * sin(omega * t), plug in the time in which the block went .04m and you will have the V(at that time and position)..**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 130%;">iii. Find its acceleration at 4 cm from the equilibrium position.

**Professor's Note: Good job on this one, solve always for "wt" and if you need you can go for "t" later**
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">a(t)=-omega^2 * X(m) * cos(omega * t) or a(t)= -omega^2 * X(m) * x(t), and we are given x(t)=.04m**

I did visit the page. I will again --- Prof !

Fourth Week 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?

First start by drawing a free body diagram... There is a mass hanging right in the middle of the string. You are given the length of the string being in equal parts as 3L/4. So, we know that the sum of the forces in the system is 2*Tension*sin(theta).

Tension can be replaced as mg, since we are solving for the mass in si units kg. State the given: length of the string is 3L/4 and they are equally divided into L/2 on both sides. mass per unit length string is said to be 8g/m convert to kg =.008kg/m you are given the wave speed =60 m/s sin (theta) can be replaced by The angle can be found using trigonometry

sin(theta)= (sqrt(7)L/8)/(3L/8) u are using the following equation

V=sqrt(Force/mass per unit length) V=sqrt((2*mg* (sqrt(7)L/8)/(3L/8))/ mass per unit length) based on the given you can now solve for the mass...Good Luck!!:)

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