Recent content by cortozld
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Find the square root of: 4+4(sqrt3)i
Woo hoo! Thanks for the help- cortozld
- Post #3
- Forum: Precalculus Mathematics Homework Help
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Find the square root of: 4+4(sqrt3)i
Homework Statement find the square root of: 4+4(sqrt3)i put the answer in a+bi form Homework Equations finding the nth roots: √(r(cos(theta)+isin(theta)))=√r(cos(theta+2(pi)(k)/(n))+isin(theta+2(pi)(k)/(n) where k=0,1 The Attempt at a Solution first i converted my...- cortozld
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- Root Square Square root
- Replies: 3
- Forum: Precalculus Mathematics Homework Help
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Solving a 2m Inclined Plane Problem with Negligible Friction
So that means i have 2gh=v^2(1+.4). Then divide by (1+.4): 2gh/(1+.4)=v^2 and vinally solve for v. which is v=sqrt(2*9.81*2)/(1+.4) so v=5.3 m/s! Thanks a lot :)- cortozld
- Post #7
- Forum: Introductory Physics Homework Help
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Solving a 2m Inclined Plane Problem with Negligible Friction
From gh=.5v^2+.5(.4)v^2 I multiplied both sides by 2 to get rid of the .5: 2gh=v^2+.4v^2. Then I divided by .4 and added the v^2s together: 2gh/.4=2v^2. Divided both sides by 2 again to get: gh/.4=v^2.- cortozld
- Post #5
- Forum: Introductory Physics Homework Help
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Solving a 2m Inclined Plane Problem with Negligible Friction
Yes, your second statement was correct. As for the problem not being specific it's my teacher not me so talk to him :). As for the answer apparently mine is wrong. So if you have any ideas about something I've missed I'd be glad to hear it- cortozld
- Post #3
- Forum: Introductory Physics Homework Help
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Solving a 2m Inclined Plane Problem with Negligible Friction
Homework Statement A solid ball is at the top of a 2 meter inclined plane. Assuming friction is negligible, what is its linear speed (m/s) as it reaches the bottom? Homework Equations PE=mgh KE=.5mv^2 RKE=.5Iw^2 I of ball=.4mr^2 w=v/r The Attempt at a Solution Pretty much what...- cortozld
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- Friction Inclined Inclined plane Plane
- Replies: 6
- Forum: Introductory Physics Homework Help
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Orbital Velocity of Equatorial Satellite at 352,000 km
so my equation will be T^2 \propto 358370000^3? I am getting a really low answer and according to my teacher it is wrong- cortozld
- Post #5
- Forum: Introductory Physics Homework Help
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Orbital Velocity of Equatorial Satellite at 352,000 km
so kepler's 3rd law is T1^2=R1^3 T2^2 R2^3 so T2 is the period, R1 is radius of earth, and R2 is radius of my orbit? what is T1 then?- cortozld
- Post #3
- Forum: Introductory Physics Homework Help
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Orbital Velocity of Equatorial Satellite at 352,000 km
Homework Statement An Earth's satelite is in equatorial orbit at 352,000 km above earth. What is the orbital velocity (m/s) of the satelite (4 sig figs) Homework Equations g1d1^2=g2d2^2 to find gravity at the height of the satellite The Attempt at a Solution I don't really know...- cortozld
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- Orbit Satelite
- Replies: 4
- Forum: Introductory Physics Homework Help
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Satelite orbiting earth in a circular path
that was an accidental typo, my bad. I believe its my teachers website, he's been having lots of problems with it since the year started. So unless anyone has an idea about what I'm doing wrong I'll just have to talk to him on Mon.- cortozld
- Post #3
- Forum: Introductory Physics Homework Help
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Satelite orbiting earth in a circular path
Homework Statement What is the orbiting speed (m/s) of a satelite orbiting Earth in a circular path 1000 km above Earth's surface? (Use 4 sig figs) Homework Equations universal constant of 6.67E-11 r of Earth 6.37E6 m (given by teacher) mass of Earth 6E24 kg (given by teacher)...- cortozld
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- Circular Earth Path Satelite
- Replies: 2
- Forum: Introductory Physics Homework Help