Problem Set 13: | 9
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19 | 23 | 31 | 33
| 35 | 37 | 10:1 | Go up

** - by David Harms, 2002**

**9. A concrete highway is built of slabs 14 m long (20Cº).
How wide should the expansion cracks be (at 20Cº) between the slabs to prevent
buckling if the temperature range is -30Cº to +50Cº?**

I know I use the formula:

DL = aL_{o}DT^{ }DL = ? (Change in length of the slabs. We are solving for this.)^{ }a = 12e-6 (Coefficient of expansion. I looked it up on page 388.)^{ }L_{o }= 14 m (Initial length of slabs.)^{ }DT_{ }= 30 Cº(50Cº - 20Cº) (Change in temperaure. You only care about the hottest number since you are dealing with expansion.)I then had the formula:

^{ }DL = 12e-6 *14 m * 30Cº^{ }DL = .00504 m^{ }DL = .50 cmThe space between the slabs must be the same as the expansion of the the slabs. Therefore .50 cm is the correct answer.

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**17. A quartz sphere is 14.5 cm in diameter. What
will its change in volume if it is heated from 30Cº to 200Cº? **

I had the formula:

DV = gV_{o}DT

DV = ? (Volumetric change in the sphere. We are solving for this.)

g = 1e -6 (Cº)^{-1}(Coefficient of volumetric expansion. It is on page 388 listed as b.)

V_{o }= (4/3) * p(.145 m/2)^{3}(Initial volume. You must take the diameter and divide by to to get the radius. Then you must stick it in the formula for volume of a sphere: (4/3)pr^{3 }V_{o }= .001569 m^{3}

DT = 170Cº (Change in temperature. 200Cº - 30Cº.)I then had the formula:

DV = 1e-6*.001569m^{3}*170Cº

DV = 2.7 e-7 m^{3}*1e6 (Conversion from m^{3}to cm^{3}.)

DV = .27 cm^{3}The answer is therefore .27 cm

^{3}.

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**19. If the fluid is contained in a long, narrow vessel so
it can expand in essentially one direction only, show that the effective coefficient
of linear expansion a is approximately equal to the coefficient
of volume expansion b (g in
the IB packet.)? **

Since it can only expand in one direction, is is really linear expansion, not volumetric expansion. Therefore the two coefficients would be the same.

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**23. A 23.4 kg solid aluminum wheel of radius 0.52 m is
rotating about its axel in frictionless bearings with an angular velocity w
= 32.8 rad/s. If its temperature is now raised from 20.0Cº to 75.0Cº,
what is the fractional change in w?**

To solve this problem we must find I (Moment of Inertia) before it heats up and after it heats up.

I have the formula:

I_{1}= Sm_{i}r_{i}^{2 }I_{1}= ? (Moment of Inertia. We are solving for this.)

S = 1/2 (looked it up on page 22).

m = 23.4 kg (mass)

r = .52 m (radius)I then had the formula:

I_{1}= (1/2)*23.4kg * .52^{2}

I_{1}= 3.16368 kg m^{2}

I then had to find the expansion of the radius using:

DL = aL_{o}DT

DL = ?

L_{o}= .52 m

a = 25e-6Cº^{-1}

DT = 55Cº

I then had:

DL = 25e-6Cº^{-1}* .52 m * 55Cº

DL = 7.14 e -4 mI then had to find I

_{2}:

I_{2}= Sm_{i}r_{i}^{2 }I_{2}= ?

S = 1/2

m = 23.4 kg

r = .520715

Then I had:

I_{2}= .5 * 23.4 *.520715^{2 }I_{2}= 3.172386 kg m^{2}

I also know I have the formula:

I_{1}w_{1}= I_{2}w_{2}

w_{1}= 32.8 rad/s (Angular velocity.)

I_{1}= 3.16368 (Found at beginning of problem.)

w_{2}= ? (We are solving for this.)

I_{2}= 3.172886 (Just solved for.)I then had:

w_{2}= 32.7048 r/s

To find the percentage difference I took

(w_{1-}w_{2})/w_{1}

( 32.8-32.7028)/32.8Which equals the answer:

.3%

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**31. If 3.0 m ^{3} of a gas initially as STP is
placed under a pressure of 4 atm, the temperature of the gas rises to
38.0Cº. What is the volume? ?**

I know I had to use the equation:

(P_{1}V_{1})/T_{1}= (P_{2}V_{2})/T_{2}

P_{1}= 1.013e5 Pa (1 atm* 1.013e5 Pa/atm)

V_{1}= 3 m^{3}

T_{1}= 273 k (Number of Kelvins at STP)

P_{2}= 405200 Pa (4 atm * 1.013e5 Pa/atm)

V_{2}= ? (We are solving for this)

T_{3}= 311 k (273 k +38)

I then solved this as got:

V_{2}= .85 cm^{3}

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