dr-zorro
dr-zorro
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L = m v r = 0.8 kg*3.6 m/s * 1.6 m = 4.61 kgm^2/s The centripetal force for the circular motion is delivered by the tension in the string: T = m v^2/r = 0.8 kg * (3.6 m/s)^2 / 1.6m = 6.48N ( you used T-mg, but the mass is on a table so the …
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If the position vector is r(t) = ( -11 + t - 3t^3 , 17 + 4 t - 9 t^2 ) then the velocity vector follows from differentiation v( t) = ( 1 - 9 t^2 , 4 - 18 t) and the acceleration vector follows from one more differentiation a(t) = ( -…
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since cos(x) dx = d ( sin(x) ), upon substituting sin(x) = u your integral is int ( 1/sqrt( 1+u^2) du ) u= 0..1 Now substitute u = sinh(v), then du = cosh(v) dv, sqrt(1+u^2) = cosh(v), so the integral just becomes int ( dv) , v = 0..arsinh(1…
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since cos(x) dx = d ( sin(x) ), upon substituting sin(x) = u your integral is int ( 1/sqrt( 1+u^2) du ) u= 0..1 Now substitute u = sinh(v), then du = cosh(v) dv, sqrt(1+u^2) = cosh(v), so the integral just becomes int ( dv) , v = 0..arsinh(1…
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If the position vector is r(t) = ( -11 + t - 3t^3 , 17 + 4 t - 9 t^2 ) then the velocity vector follows from differentiation v( t) = ( 1 - 9 t^2 , 4 - 18 t) and the acceleration vector follows from one more differentiation a(t) = ( -…
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L = m v r = 0.8 kg*3.6 m/s * 1.6 m = 4.61 kgm^2/s The centripetal force for the circular motion is delivered by the tension in the string: T = m v^2/r = 0.8 kg * (3.6 m/s)^2 / 1.6m = 6.48N ( you used T-mg, but the mass is on a table so the …