Download Practice Problems for Test III

Practice Problems for Test III
Phy2020 Test will be on April 11, 2008 Chapters 12-19 (no 15)
g=9.8 m/s2 ; 1 Hz = 1 cycle/second ; 1000 g = 1 kg
3 sig figs for all answers (-1 point each omission)
1. How much lift force is there on a object with volume 500 cm3 that is
immersed entirely in a fluid with density 0.89 g/cm3? (get the units right –it’s
a force.)
Soln: 500 cm3 of fluid displaced is 500 x 0.89 g = 445 g. By Archimedes’
principle, this give a Flift=mass displaced x g = 4.361 N.
2. A gas that has no (or not much) interaction between the atoms/molecules is
called ideal. An example is Helium (chemical symbol He) gas. If a 1000 cm3
of He gas in a fixed volume has its temperature = 80 oF and its pressure = 1
bar, what is the pressure going to be if the temperature is changed to 160 oF?
(3 sig figs please, units of bar (=1 atmosphere of pressure, or 15 lbs/in2)
PV=nRT if the volume is fixed, then P is proportional to T, as long as (as we
discussed in class) T is in units of K. Convert 80 oF to oC: (80-32)/1.8=26.67 oC.
Now convert 160 oF to oC: (160-32)/1.8 = 71.11 oC. Now convert the two
temperatures in oC to K (you could also have gone directly from oF to K):
26.67 oC = 273 + 26.67 K = 299.67 K, while 71.11 oC = 344.11 K, so since the
temperature increases by a factor of 344.11/299.67 = 1.148, the pressure
increases by the same factor, so Pnew=1.148 bar.
3. A spring scale stretches 5 cm when a 3 kg mass is hung from it.
If the mass is released from its stretched position, what is the frequency in
cycles/sec, of the oscillatory motion that results?
Soln: As was done for the practice problem similar to this, first find the spring
constant
F=-kx, so the force due to the 3 kg mass is 3 kg*9.8 m/s2 = 29.4 N, for a stretch (x) of
0.05 m, so k=29.4 N/0.05 m = 588 N/m
Assuming the mass of the spring is negligible compared to the 3 kg mass, f=/2,
where =sqrt(k/m)=sqrt(588 N/m/3 kg) = sqrt (196 kgm/s2/mkg) so =14 rad/sec
so f=2.23 cycles/s or 2.23 Hz.
4. If the maxima in a water wave are 4 m apart, and an observer notices at a
certain point that the water crests, troughs, and crests again in 5 seconds,
what is the velocity of the water wave, in m/s?
Solution: the wavelength λ is 4 m, and the period, T, (the time it takes for one cycle
of the water wave) is 5 seconds. f=1/T and velocity of wave in medium = λf,
so the velocity of the water wave is 4 m * 1/(5 s) = 0.8 m/s
5. How much louder (by what factor is the sound intensity larger) is 105 dB
sound than a 90 dB sound?
Loudness (sound intensity) is described by the decibel scale,
β (in dB) = 10 log(I/I0), where I0 is the reference sound intensity 10-12 W/m2
Therefore, 105 dB has an I given by 105 = 10 log(I/I0)
And 90 dB has an I given by 90 = 10 log(I/I0) or 9=log(I/I0) or 9=logI – logI0
Or 9= logI –(-12) or logI = -3
So I=10-3 W/m2 and I (from the 105 dB) can be found from 10.5 = logI + 12
So I=0.0316 W/m2, to the ratio of loudness is .0316/10-3 = 31.6
6. A source of sound (vsound=343 m/s) is receding from an observer at 100 m/s.
If the source is emitting 1000 Hz sound, what frequency (in Hz) does the
observer detect?
Use the Doppler formula for the source moving away from the observer:
f’ = f/(1 + vsource/vsound) = 1000 Hz/(1+100/343) = 774 Hz.
A
B
R1
R2
R3
7a. (1 point). If three resistors, R1= 10 Ω, R2= 25 Ω, and R3= 15 Ω are hooked in
series between point A and point B, what is the resultant total resistance of these
three resistors?
Soln: Rtotal in a series resistor arrangement is just the simple sum of the values of
the resistors = 10+25+15 Ω = 50 Ω
b. (1 point). If a current of 2 A leaves point A in 7a above (the series resistor
arrangement), then the power loss across the sum of all 3 resistors is I2Rtotal, where
Rtotal is your answer to 7a. What is the power loss caused by the current I flowing
through resistor R1? Give the answer in units of Watts.
Power loss = I2R1 = 4 A2 x 10 Ω = 40.0 W
8. The heat capacity of water is about 1 cal/goC, i. e. it takes 1 Calorie of
energy (=4.184 J) to warm one gram of water by 1 oC. An inventor from an
unnamed university in Florida (not U of F!) has ‘discovered’ a new source of
almost inexhaustible energy (he/she thinks) which will also counteract global
warming (and create perpetual motion, but ignore that part.) This person’s
idea is to cool all of the oceans of the world off by 1 oC and use the energy
obtained to run the world’s demand for electrical energy. If the mass of the
water on the Earth is 1.40 × 1021 kilograms, how much energy in J would
cooling this mass of water by 1 oC provide? (Please be careful of the units –
only 1 point if you get the number right but the exponent of 10 wrong.)
1.40 × 1021 kilograms = 1.40 × 1024 grams
so cooling this mass of water by 1 degree gives 1.40 × 1024 cal = 5.86 x 1024 J.