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ReMa

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Hi There,

I have a handful of questions I need help with and figured compiling them in one thread was the best option. For the most part, i've added these because I just don't know where to start and can't figure it out at all. Here we go!

A bar of copper with length 2.635m and a bar of Aluminum with length 2.628m are sitting at room temperature, T1 = 25C. At what temperature, T2, will the two have the same length?

The coefficient of thermal expansion of copper is 1.60x10-7 K

Linear Thermal Expansion: [tex]\Delta[/tex]L=[tex]\alpha[/tex]Lo[tex]\Delta[/tex]T

Lo = initial Length

The given choice of answers are in C, so I figured I had to convert the thermal expansion coefficients to C

K = C + 273

I figured the expansion would be Lo + [tex]\Delta[/tex]L

I set that eq. for the Al and Cu bars equal, subbing in the [tex]\Delta[/tex]L with the equation above. My answer is off though, and i'm not sure if it's a matter of because I didn't convert my per Kelvin to per Celsius or what. I think it's an algebra thing but maybe i'm way off.

The sound level at a point X is 14db below the sound level at a point 1.0m from a point source. The distance from the source to point X is:

a) 2.0m

b) 25m

c) 5.0m

d) 20.2m

e) 4.0m

Decibels = 10 log (I/Io)

Io = threshold of hearing = 1x10

I don't know where to start with this one!

A heat conducting rod, 1.60m long, is made of an Al section, 0.90m long, and a Cu section, 0.70m long. Both sections have a cross sectional area of 0.0004m2. The Al end and the copper end are maintained at temperatures of 30C and 170C respectively. The thermal conductivieis of Al and Cu are: kal = 205 W/m K and kcu = 385 W/m K. The rate at which heat is conducted is:

a) 10W

b) 9W

c) 7.9W

d) 12W

E) 11W

Q = (kA[tex]\Delta[/tex]T)t/L

I couldn't figure out how to fit the t (time) into this at all, so perhaps i'm using the wrong equation?

k values were all given

A = 0.0004m2

[tex]\Delta[/tex]T = 170-30 = 140C

Otherwise, I am also confused with this one. Do I need to find a seperate Q for both materials (Al and Cu)? And if so, what happens to that t (time) value?

A proton is fixed in place. A second proton is released from rest at a point 1.0cm away What will be the velocity of the second proton when it is a very large distance from the first?

Not sure, Wnc = 1/2 mvf

Proton at rest = vo = 0m/s

Proton mass = 1.67 x 10

A bus is moving at 37.00m/s towards a wall. The sound from the bus has an original wavelength of 0.1500m. The sound from the bus reflects off the wall. What frequency sound does

Moving Observer: fo = fs (1 + vo/v)

v = [tex]\lambda[/tex]f

Is this doppler effect??

I'm assuming the wavelength will double given that it reflects off a wall?

So, [tex]\lambda[/tex] = 2(0.1500m) = 0.300m

vo = 37.00m/s

Since v = [tex]\lambda[/tex]f

vi = [tex]\lambda[/tex]fi

f = v / [tex]\lambda[/tex]

= (343m/s)(0.1500) = 2286.667hz

Subbing into equation:

fo = (2286.667hz)(1 + 37m/s / 343m/s) = 2533hz

Is that correct??? I don't think it is because doing that didn't account for the reflection off the wall - unless I misunderstand how that works?

---------

Thanks for any help!!

I have a handful of questions I need help with and figured compiling them in one thread was the best option. For the most part, i've added these because I just don't know where to start and can't figure it out at all. Here we go!

__QUESTION ONE:__## Homework Statement

A bar of copper with length 2.635m and a bar of Aluminum with length 2.628m are sitting at room temperature, T1 = 25C. At what temperature, T2, will the two have the same length?

The coefficient of thermal expansion of copper is 1.60x10-7 K

^{-1}and of Aluminum is 2.25x10-7 K^{-1}.## Homework Equations

Linear Thermal Expansion: [tex]\Delta[/tex]L=[tex]\alpha[/tex]Lo[tex]\Delta[/tex]T

Lo = initial Length

## The Attempt at a Solution

The given choice of answers are in C, so I figured I had to convert the thermal expansion coefficients to C

^{-1}but wasn't quite sure how to do that.K = C + 273

I figured the expansion would be Lo + [tex]\Delta[/tex]L

I set that eq. for the Al and Cu bars equal, subbing in the [tex]\Delta[/tex]L with the equation above. My answer is off though, and i'm not sure if it's a matter of because I didn't convert my per Kelvin to per Celsius or what. I think it's an algebra thing but maybe i'm way off.

__QUESTION TWO:__## Homework Statement

The sound level at a point X is 14db below the sound level at a point 1.0m from a point source. The distance from the source to point X is:

a) 2.0m

b) 25m

c) 5.0m

d) 20.2m

e) 4.0m

## Homework Equations

Decibels = 10 log (I/Io)

Io = threshold of hearing = 1x10

^{-12}W/m2## The Attempt at a Solution

I don't know where to start with this one!

__QUESTION THREE:__## Homework Statement

A heat conducting rod, 1.60m long, is made of an Al section, 0.90m long, and a Cu section, 0.70m long. Both sections have a cross sectional area of 0.0004m2. The Al end and the copper end are maintained at temperatures of 30C and 170C respectively. The thermal conductivieis of Al and Cu are: kal = 205 W/m K and kcu = 385 W/m K. The rate at which heat is conducted is:

a) 10W

b) 9W

c) 7.9W

d) 12W

E) 11W

## Homework Equations

Q = (kA[tex]\Delta[/tex]T)t/L

## The Attempt at a Solution

I couldn't figure out how to fit the t (time) into this at all, so perhaps i'm using the wrong equation?

k values were all given

A = 0.0004m2

[tex]\Delta[/tex]T = 170-30 = 140C

Otherwise, I am also confused with this one. Do I need to find a seperate Q for both materials (Al and Cu)? And if so, what happens to that t (time) value?

__QUESTION FOUR:__## Homework Statement

A proton is fixed in place. A second proton is released from rest at a point 1.0cm away What will be the velocity of the second proton when it is a very large distance from the first?

## Homework Equations

Not sure, Wnc = 1/2 mvf

^{2}???## The Attempt at a Solution

Proton at rest = vo = 0m/s

Proton mass = 1.67 x 10

^{-27}kg__QUESTION FIVE:__## Homework Statement

A bus is moving at 37.00m/s towards a wall. The sound from the bus has an original wavelength of 0.1500m. The sound from the bus reflects off the wall. What frequency sound does

*an observer on the moving bus*hear from the reflection??## Homework Equations

Moving Observer: fo = fs (1 + vo/v)

v = [tex]\lambda[/tex]f

## The Attempt at a Solution

Is this doppler effect??

I'm assuming the wavelength will double given that it reflects off a wall?

So, [tex]\lambda[/tex] = 2(0.1500m) = 0.300m

vo = 37.00m/s

Since v = [tex]\lambda[/tex]f

vi = [tex]\lambda[/tex]fi

f = v / [tex]\lambda[/tex]

= (343m/s)(0.1500) = 2286.667hz

Subbing into equation:

fo = (2286.667hz)(1 + 37m/s / 343m/s) = 2533hz

Is that correct??? I don't think it is because doing that didn't account for the reflection off the wall - unless I misunderstand how that works?

---------

Thanks for any help!!

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