Recent content by Kotune
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K
Gymnast swinging on bar with 1.6kN force at lowest point
thanks I got it ^_^- Kotune
- Post #20
- Forum: Introductory Physics Homework Help
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K
Gymnast swinging on bar with 1.6kN force at lowest point
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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K
Gymnast swinging on bar with 1.6kN force at lowest point
? then is v = 7.85?- Kotune
- Post #16
- Forum: Introductory Physics Homework Help
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K
Gymnast swinging on bar with 1.6kN force at lowest point
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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K
Gymnast swinging on bar with 1.6kN force at lowest point
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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K
Gymnast swinging on bar with 1.6kN force at lowest point
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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K
Gymnast swinging on bar with 1.6kN force at lowest point
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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K
Gymnast swinging on bar with 1.6kN force at lowest point
Hello! :) I get ai) now ty! I got 4.8ms-1 :)- Kotune
- Post #3
- Forum: Introductory Physics Homework Help
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K
Gymnast swinging on bar with 1.6kN force at lowest point
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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K
Landing Ramp Distance Required for Stunt Car Launch
got it, cheers buddy!- Kotune
- Post #12
- Forum: Introductory Physics Homework Help
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K
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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K
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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K
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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K
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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K
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