Recent content by Kotune
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Solving Circular Motion Homework - 1.6kN Force to Hang on
thanks I got it ^_^- Kotune
- Post #20
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
I don't get it. :frown: The equation would then be .5 x (40) x (v+1)2 = (40) x (9.8) x (2.4)?- Kotune
- Post #18
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
? then is v = 7.85?- Kotune
- Post #16
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
Sorry dumb question but why is h the diameter? xD .5(40)v2=40 x 9.8 x 2.4 v = 6.86 Is tension = centripetal force + weight? = then mv2/r + m x 9.8 = 40 x 6.85 2 / 2.4 + (40 x 9.8) = 1176N which is wrong ><- Kotune
- Post #14
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
Does that mean h is 1? .5mv2 = mgh .5(40)v2 = 40 x 9.8 x 1 then the speed at the lowest point is 4.4? Edit: wait no since its diameter then the h is 2.4?- Kotune
- Post #12
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
for the mgh. Mass is still 40kg, gravity is still 9.8, that means the height has to change? But it can't be 0 ><.- Kotune
- Post #10
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
I'm confused to as how to find the speed at the bottom because isn't the speed constant?- Kotune
- Post #7
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
Hello! :) I get ai) now ty! I got 4.8ms-1 :)- Kotune
- Post #3
- Forum: Introductory Physics Homework Help
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Solving Circular Motion Homework - 1.6kN Force to Hang on
Homework Statement A 40kg gymnast swings in a vertical circle on a bar. Her centre of mass is 1.20m from the bar. At the highest point her centre of mass is moving at 1.0ms^-1) ai) How fast is she moving when her centre of mass is level with the bar? aii) How much force does she have to...- Kotune
- Thread
- Circular Circular motion Motion
- Replies: 19
- Forum: Introductory Physics Homework Help
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Landing Ramp Distance Required for Stunt Car Launch
got it, cheers buddy!- Kotune
- Post #12
- Forum: Introductory Physics Homework Help
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Landing Ramp Distance Required for Stunt Car Launch
.7 x 39.39 = 27.573 Then do I multiply this number by 2?- Kotune
- Post #10
- Forum: Introductory Physics Homework Help
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Landing Ramp Distance Required for Stunt Car Launch
v = u + at? 0 = 6.95 +9.8t -6.95 / 9.8 = t 0.7 = t 0.7s- Kotune
- Post #8
- Forum: Introductory Physics Homework Help
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How fast did you throw the ball in order to get it into the basket?
Thanks. How would you arrange it for u? √1.5 - (3tan(55)) - (9.8x3^2) x (2(cos55)) = u?- Kotune
- Post #7
- Forum: Introductory Physics Homework Help
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Landing Ramp Distance Required for Stunt Car Launch
x = 40cos10 = 39.39 y = 40sin10 = 6.95 I get that part but I'm not sure how that helps me achieve the conclusion? ><- Kotune
- Post #6
- Forum: Introductory Physics Homework Help
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How fast did you throw the ball in order to get it into the basket?
u = 1.5cos 55 = 0.86 y = 3tan(55) - (9.8x3^2) / 2 (0.86cos55)^2 = -172... Thats not correct.- Kotune
- Post #5
- Forum: Introductory Physics Homework Help