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In the circumstance described the component of F perpendicular to the rod is 1/2 F In the second case the component is F ie F * x = 1/2 F * 40

**cm**-> x = 20**cm**1 Answers · Science & Mathematics · 04/03/2021

torque = Fd = 40 N •0.2 m = 8 Nm

2 Answers · Science & Mathematics · 04/03/2021

At ½ the maximum speed, it has ¼ of the maximum KE, which means it has 3/4 of the maxim PE, which means x = 4.37

**cm*** √(3/4) = 3.78**cm**1 Answers · Science & Mathematics · 04/03/2021

Since it begins and ends "at rest," KE is not involved and the mass's GPE gets converted into spring SPE: mgh = ½kh² m/k = h / 2g = 0.0616m / 2*9.81m/s² = 0.00314 s² T = 2π√(m/k...

2 Answers · Science & Mathematics · 04/03/2021

max acceleration a = Aω² ω² = a/A = 10m/s² / 0.741m = 13.5 rad/s² and also ω² = k/m, so 13.5 rad/s² = k / 5.8kg making k = 78.3 kg/s² (or N/m) total energy TME = ½kA² = ½ * 78.3N/m * (0...

2 Answers · Science & Mathematics · 04/03/2021

...3 *(Pi*√3) = 4*Pi = 0.2 Pi = 0.2/4 = 0.05 = 1/20 or 5

**cm**/s <<< Ps = Pi *√3 = 0...1 Answers · Science & Mathematics · 03/03/2021

radius = 0.25

**cm**= 0.0025 m area = πr² = 0.00001964 m² R = (1.68e-8)(5000)/(0...1 Answers · Science & Mathematics · 03/03/2021

You must be consistent with units. Either use dm for everything which gives a result in Litres and kg or work in m^3 which gives a result in tonnes. Volume displaced is 4*6 * 0.04 = 0.96 m^3 and with a mass of 1 tonne / m^3 this is...

1 Answers · Science & Mathematics · 02/03/2021

max acceleration a = Aω² = A(2πf)² = 4π²Af² so a = 10 * 9.81m/s² = 4π² * 0.095m * f² solves to f = 5.11 s⁻¹ = 5.11 Hz

1 Answers · Science & Mathematics · 03/03/2021

a) PE becomes KE: ½kx² = ½mv² ½ * 600N/m * (0.0617m)² = ½ * 1.69kg * V² solves to V = 1.16 m/s b) PE becomes KE and friction work: ...

1 Answers · Science & Mathematics · 02/03/2021