Recent content by mccarthyp64
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Energy released in the fission of one uranium-235 nucleus? (extra neutron)
1. How much energy is released during the fission of one uranium-235 nucleus? 2. e=mc2 lol :smile: 3. I thought of putting them together but then how to separate them? 1.67495×10-27+3.903 05×10–25×c2=1.7539207313048996244767495×10-7 By the way, the answer is 4.73×10-11- mccarthyp64
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- Energy Energy released Fission Neutron Nucleus
- Replies: 1
- Forum: Introductory Physics Homework Help
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Calculating Velocity in Circular Motion
Ep=mgh=Ek=1/2mv2 1.176=0.4*9.8*0.3=1.176=0.5(0.4)(v)2 (1.176/0.5)/0.4=v2=5.88 √(5.88)=2.42=ANS! YAY thanks gneill :)- mccarthyp64
- Post #7
- Forum: Introductory Physics Homework Help
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Calculating Velocity in Circular Motion
for q 25 velocity is the answer and in 26 I need the work done- mccarthyp64
- Post #5
- Forum: Introductory Physics Homework Help
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Calculating Velocity in Circular Motion
There is kinetic and potential gravitational energy so Ek=1/2mv2 but that needs velocity so: Ep=mgh but that only gives horizontal velocity whereas, I need all velocity.- mccarthyp64
- Post #3
- Forum: Introductory Physics Homework Help
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Calculating Velocity in Circular Motion
1. All in the picture 2.v2=u2+2as 3. I have no idea where to start, I went through all the equations but i don't have time or velocity to sub in :( It also has circular motion which may or may not introduce pi. I didn't think this would be relevant but anyway.- mccarthyp64
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- Circular Path Velocity
- Replies: 6
- Forum: Introductory Physics Homework Help
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Archimedes' Dilemma: The Misunderstood Concept of Pi = 4
Problem being that repeating gives you an octagon http://upload.wikimedia.org/wikipedia/commons/6/66/Regular_octagon.svg Instead of a circle, because the side of a circle isn't straight etc. P.S. I only figured this out 2 days ago- mccarthyp64
- Post #50
- Forum: General Math