StarSecret Posted July 8, 2011 Report Share Posted July 8, 2011 A 0.50 kg cue ball makes a glancing blow to a stationary 0.50 kg billiard ball. After the collision the cue ball deflects with a speed of 1.2 m/s at an angle of 30.0º from its original path. Calculate the original speed of the cue ball if the billiard ball ends up travelling at 1.6 m/s.thanks Reply Link to post Share on other sites More sharing options...
Drake Glau Posted July 8, 2011 Author Report Share Posted July 8, 2011 Hmmm either this is vectors or the angle is there to screw with you...because I feel like it's not needed... You have to conserve momentum which is p=mv... so 0.50*1.2=0.6 and 0.50*1.6=0.8 So the initial momentum had to be 1.4 and by rearranging p=mv you can get v=p/m or v=1.4/0.5 which is 2.8m/s... But I'm not comfortable leaving the angle out Reply Link to post Share on other sites More sharing options...
StarSecret Posted July 8, 2011 Report Share Posted July 8, 2011 Hmmm either this is vectors or the angle is there to screw with you...because I feel like it's not needed... You have to conserve momentum which is p=mv... so 0.50*1.2=0.6 and 0.50*1.6=0.8 So the initial momentum had to be 1.4 and by rearranging p=mv you can get v=p/m or v=1.4/0.5 which is 2.8m/s... But I'm not comfortable leaving the angle out thanks for the help Reply Link to post Share on other sites More sharing options...
Guest kenshi64 Posted July 8, 2011 Report Share Posted July 8, 2011 It was already answered, sorry =/ Apparently he asked in two places... Sorry! still getting used to the rules, won't take long, I'm here almost daily!! Reply Link to post Share on other sites More sharing options...
genepeer Posted July 8, 2011 Report Share Posted July 8, 2011 Does the mass of a pendulum affect its period? For small amplitude of swings, the formula for period (T) is L = length of string/pendulum g = gravity If I changed the mass, would it affect the formula? Reply Link to post Share on other sites More sharing options...
Guest KAPOWW!! Posted September 7, 2011 Report Share Posted September 7, 2011 Doubt: When the textbook says that during beta decay the neutron changes to proton and electron(Beta negative, I assume)So is my assumption right? Reply Link to post Share on other sites More sharing options...
Guest PoisonousPumpkin Posted September 8, 2011 Report Share Posted September 8, 2011 Could you help me please defining the luminosity of a star and stating one factor that determines the luminosity of a star? Reply Link to post Share on other sites More sharing options...
Procrastination Posted September 8, 2011 Report Share Posted September 8, 2011 Easy question. The luminosity of a star is the total power emitted. There are many factors that determine the luminosity of a star:Type of starTemperatureMassSurface areaRadiusCompositionThere you go. 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted September 9, 2011 Report Share Posted September 9, 2011 Doubt: When the textbook says that during beta decay the neutron changes to proton and electron(Beta negative, I assume) So is my assumption right? eh... during beta decay the nucleus releases an antineutrino. it loses an electron, so it kinda gets one more proton. in the case of a neutron, yes it is right. you must have seen this: in that case, yes it becomes a proton because the nucleus now gains one proton, and since it initially had 0 proton, now it becomes a "proton". but in a beta decay, the nucleus doesn't always change into proton. if the decaying nucleus is not only a neutron, they will become the next element in the periodic table as it now has one more proton. example: like Carbon now has 7 protons, but it's no longer Carbon. it's Nitrogen. same with the second example. your assumption is correct but remember if it's not just a neutron then it'll be a different element not just a proton. Reply Link to post Share on other sites More sharing options...
Guest -Guns of Steel- Posted October 12, 2011 Report Share Posted October 12, 2011 Hey guys can you have a look at my lab? I think there's a problem with it. Check out the file attached, the problem I'm facing is that the graph is a best-fit built of the values in the table so there's no point in the comparing the difference between the two Rs right? (R1-being resistance obtained as average of the 4 values, R2 being Gradient of graph obtained from those values itself! Reply Link to post Share on other sites More sharing options...
Drake Glau Posted October 12, 2011 Author Report Share Posted October 12, 2011 Part of me is really confused because the gradient of that line should be your resistance, another part of me is saying that your experiment had some error in it. When we did a lab over Ohm's law we ALL had perfect data where every point landed directly on the line.Because of this I would definitely compare the two values and discuss which one would be more accurate (I'm voting that the average is more accurate) and discuss any possible problems that could have lead to the different values. One that I can think of is that you only have 4 data points which I don't personally consider sufficient data (our teacher required a minimum of 10) and the two points with the greater amps are off the line enough to screw the slope up. Having more data points would of aided in keeping the slope closer to what it should be. This could also be opposite, the two lesser amp point could be the erroneous points, either way results in a greater slope though. I'm voting the average being more accurate since either pair of points could be the "false" ones. Reply Link to post Share on other sites More sharing options...
Chronofox Posted November 1, 2011 Report Share Posted November 1, 2011 Two questions have me stumped. For the question below, I have no idea where to begin. "When the sand hits the conveyer belt, its speed is zero." What does that mean? Reply Link to post Share on other sites More sharing options...
jonathan810 Posted November 6, 2011 Report Share Posted November 6, 2011 Reply Link to post Share on other sites More sharing options...
Drake Glau Posted November 6, 2011 Author Report Share Posted November 6, 2011 Well, we have speeds and are asked about kinetic energy. Ek=0.5mv2 but since the m and the 0.5 really don't change...at all...so we just need to know that the velocity is squared.The kinetic energy BEFORE collision is of course related to the speed before the collision, but it's also squared because of the relation to the kinetic energy so that would be u2. The kinetic energy after collision would be v2 but the kinetic energy loss would imply some subtraction so I'm going to go with D. Reply Link to post Share on other sites More sharing options...
jonathan810 Posted November 8, 2011 Report Share Posted November 8, 2011 Reply Link to post Share on other sites More sharing options...
Peanut Butter Jelly Posted November 8, 2011 Report Share Posted November 8, 2011 is it the area between the two axis? So it would be +8 since change is always positive Reply Link to post Share on other sites More sharing options...
jonathan810 Posted November 8, 2011 Report Share Posted November 8, 2011 ah.... shoot.....the area. cheers. Won't the ammeter and voltmeter both read 0? (answers says otherwise O.o) Reply Link to post Share on other sites More sharing options...
Peanut Butter Jelly Posted November 8, 2011 Report Share Posted November 8, 2011 ah.... shoot.....the area. cheers. Won't the ammeter and voltmeter both read 0? (answers says otherwise O.o) The voltmeter should read 6 cuz it stays the same and the ammeter should be 1 cuz Rtotal = 12, and therfore current = 2 and the current through that is 1. I could be wrong though someone correct me. Reply Link to post Share on other sites More sharing options...
jonathan810 Posted November 8, 2011 Report Share Posted November 8, 2011 the marking scheme says that it is B, current is 0..... Reply Link to post Share on other sites More sharing options...
Drake Glau Posted November 9, 2011 Author Report Share Posted November 9, 2011 ah.... shoot.....the area. cheers. Won't the ammeter and voltmeter both read 0? (answers says otherwise O.o) The voltmeter should read 6 cuz it stays the same and the ammeter should be 1 cuz Rtotal = 12, and therfore current = 2 and the current through that is 1. I could be wrong though someone correct me. V=IR I=V/R, that'd be 6/12=0.5amps, then splits to 0.25amps. Anyway, my guess would be to consider the resistance of the voltmeter. All voltmeters are made with really high resistances because they are usually placed in parallel causing the inverse resistance thing yadayada. Considering it is now in series and still has the high resistance that would make your resistance skyrocket and sadly it's in the denominator of ohm's law so your current just dropped like a rock, and then was split it two to make it worse ^My best guess, I'd probably would've just done the circuit and found out that 1amp isn't possible and then just go with B since I don't have time to waste on actually figuring it out during the actual paper 1 Reply Link to post Share on other sites More sharing options...
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