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course Phy 242
2-27
110216 physics`q001. A BB of mass .12 g is shot at 80 m/s into the space between two tiles. The tiles are separated by 15 cm.
If the BB continues bouncing back and forth between the tiles for 2 seconds, without losing any of its speed, what average force does it exert on each tile?
Knowing F_ave*`dt = m*`dv, from impulse logic
At collision
m*`dv = .12g*(80m/s*2) = .12g*160m/s???we mult vel by 2 because after contact the the velocity is +80 then after next hit is -80 for `dv = 160m/s???
= 19.2(g*m)/s
`dt = `ds/v_ave = 30cm/80m/s =.00375sec
F_ave = m*`dv/`dt = .0192((kg*m)/s)/ .00375sec =5.12N
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How would the result change if the tile separation was reduced to 5 cm?
`dt = 10cm/80m/s = .00125sec, therefore
F_ave = m*`dv/`dt = .0192((kg*m)/s)/ .00125sec =15.36N
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How would the result change if the BB was fired into a slightly irregular tile-lined 'box' approximately 5 cm on a side?
The path more than likely would not follow the same path because the bb’s path would vary depending where it hit on the irregular.
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What does this question have to do with the kinetic theory of gases?
The bb’s represent the path traveled by the molecules of the gases. Monatomic gases travel on a fixed axis that’s why when it comes to the kinetic theory for monatomic gases they account for less KE then diatomic gases. For diatomic which have rotational movement as well they follow a much more irregular path and could be thought of as the bb bouncing off of the irregular tile.
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Optional problem: If the BB loses 10% of its momentum with each collision, what average force does it exert over the first 10 round trips? General College Physics students can use estimates, as can University Physics students. However University Physics students should consider applying calculus and/or differential equations.
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`q002. A wave is observed to travel down a rubber band chain of length 2.5 meters, making seven round trips in 10 seconds. What is the propagation velocity of that wave?
propagation velocity = `lambda*freq.
Or can be viewed as pt at crest of wave moving along parallel to x axis we can then figure
7cycles(2*(2.5meters))/10sec = 3.5meters/sec
For 7cycles in 10sec, freq. = .75cycles/sec, now we can find wave length
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By sending alternate pulses down the chain on alternate sides, a standing wave is created with a single antinode, located halfway down the chain. How much time must elapse between the pulses?
It takes 1sec to complete .75cycle or approx. 1.33sec to complete a whole cycle. We would want to start the 2nd wave right as the 1st wave is right at mid cycle or the opposite end of the rubber band.
(1/2)cycle needs (1/2) of 1.33sec to complete
(1/2)*1.33sec = (2/3)sec
So pluses would need to be sent out every 2/3sec. Also we know that we are accounting for one cycle because we are the info about there being one antinode.
????I’m not sure this is correct????
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How many pulses are sent by the time the first disturbance reaches the end of the chain? What is the shape of the chain at that instant?
The pulse rate is increased until the standing wave contains two antinodes, with a node in the middle. How much time must elapse between the pulses?
I’m not sure I’m visualizing this correctly, but if I am then the chain would be flat at that instance??
The anti-nodes would have to be at the positions 1/3 and 2/3 down the chain
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How many pulses are sent by the time the first disturbance reaches the end of the chain? What is the shape of the chain at that instant?
I believe there will be 6?????
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If a complete cycle consists of two alternate pulses, then in each case, how many complete cycles of the wave correspond to its length?
????I’m having a hard time seeing what’s
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`q003. Suppose the tension in a rubber band chain is 1 Newton for every 10% change in its length. If our chain has length 2.5 meters when under a tension of 1 Newton, then what will be the tension when its length is 2.7 meters, what will be the tension at length 3 meters, and at what length will the tension be 5 Newtons?
8% change at 2.7m = .8N
20% change at 3m = 2N
200% change at 5m = 20N
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If the speed of the wave is c = sqrt( T / (m/L) ), then what is (m / L) for our chain (use also the information you obtained in the preceding question)?
For c = sqrt( T / (m/L) )
c^2 = T / (m/L)
(m/L) = T/c^2
= 1N/(3.5m/s)^2 = 1N/12.25(m/s)^2 = approx. .08kg/m
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What will be the wave speed at length 3 meters?
c = `sqrt(T/ (m/L) )
We know (m/L) which when mult by L gives us m
m = (m/L)*L = .08kg/m*2.5m =.2kg
Now we can simply sub in values for c = `sqrt(T/ (m/L) )
c = `sqrt(2N / (.2kg /3m) ) = 30m/s
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At length 3 meters, how frequently should pulses be sent in order to create the fundamental mode of vibration (the one with a single antinode)?
??????I’m not sure I’m doing this correctly any feedback would be appreiciated????
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02-27-2011