Recent content by habibclan

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    Superposition and Interference problem

    Can someone help me please?? I've got my exam tomorrow morning!
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    Superposition and Interference problem

    You're referring to the frequency equation: f= mv/2L. I tried using that but the answer comes out to 8.33 m, whereas at the back of the textbook, it is 18 cm :S. This is my working: frequency for the 8 cm one where m=3 as there are 3 antinodes: f= (3 v) / (2*0.08) frequency for unknown...
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    Superposition and Interference problem

    I have two questions from the Superposition unit that I don't get it. I'd really any help on these as I have my physics exam on monday! Thanks for any help! Homework Statement A steel wire is used to stretch a spring. An oscillating magnetic field drives the steel wire back and forth. A...
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    Elastic Collision of a gold nucleus

    My formula sheet has a weird way of writing the formula and they had m2 on there, thats why instead of having 2 in my numerator, i have 100. You guys saved me from using that formula on the exam! Thanks a lot! [Now my logical answer matches the math too! Yay!]
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    Elastic Collision of a gold nucleus

    hehe. i do get that. the only discrepency is that the formulas give a greater velocity for the gold nucleus. Therefore, making it have a higher KE. Or should I not use the formula over here and it's not a REAL elastic collision.
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    Elastic Collision of a gold nucleus

    This question is from a past midterm and the answer given in the answer key is D. I've thought of it in terms of momentum and since velocity is squared in the KE expression, larger velocity of the alpha particle gives it a greater KE. Besides this, I don't know what to conclude from this...
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    Elastic Collision of a gold nucleus

    Homework Statement A positively charged alpha particle is fired, in a direct line, at a positively charged gold nucleus: o--> O o: alpha, O: golden nucleus. Because of their mutual repulsion, the alpha particle does not actually hit the nuleus: it comes to rest, for a...
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    Kinematics of jet-ski - how far does Sam land?

    I got the answer now! Yay! Thank you so much for guiding me! I love this forum! I'm feeling more confident now and learning more things by second :D.
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    Kinematics of jet-ski - how far does Sam land?

    Thanks a lot! That solves the mystery. I assumed that the force of thrust was only present while on the ramp. So now, I have an acceleration in both x and y directions and I can use a similar method to solve for the horizontal distance.
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    Kinematics of jet-ski - how far does Sam land?

    For the acceleration, you mean? The overall answer for some reason isn't turning out to be correct. I checked my calculations many times. In the beginning, I made a few error but then I corrected those, but the answer I'm supposed to be getting is 110 m, not 84.4 m :(. I hate textbook questions...
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    Net Work

    I was thinking of the work-kinetic energy theorem were Wnet= delta K. But maybe I should always stick to the equation: Ki + Ugi + Wext= Kf + Ugf + Delta E (thermal), and plug in things to see where the conversion is occuring. My textbook gives too many formulas which creates a lot of confusion!
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    Net Work

    Sorry :$. It's just that I'm not really good with the concept of work. I read the textbook and did all the recommended questions but this was from a practice midterm. I'll try not to post as much now and try to think them through. But I only post questions if I try them at least 3-4 times and...
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    Energy Conservation.

    Nice!! Thank you so much!!!!
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