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  1. F=k ((q1*q2)/r^2) 9.0N=(9*10^9 N m^2 C^-2)*((-0.0005 C...

    2 Answers · Science & Mathematics · 15/01/2018

  2. ...diagram: Fc←●→Fb Fnet = Fb − Fc Fnet = k q₁ q₁ / r² − k q₁ q₂ / (2r)²...

    2 Answers · Science & Mathematics · 14/01/2018

  3. ... da (deca) = x10 (rarely used) h (hecto) = x100 (rarely used) k (kilo) = x1000 = e3 M (Mega) = x1000000 = e6 G (Giga) = x1000000000...

    2 Answers · Science & Mathematics · 26/12/2017

  4. ... GPE at top = PE of spring at bottom m g h = 1/2 k x^2 h = (d + 0.25) sin30 = 0.5 d + 0.125 so 0.5 (9.8) (0.5 d + 0.125) = 1/2...

    1 Answers · Science & Mathematics · 17/12/2017

  5. While she is sliding down the slope, the net force is the component of her weight that is parallel to the slope minus the kinetic friction force. Force parallel = m * 9.8 * sin 31 Ff = 0.13 * m * 9.8 * cos 31 a = 9.8 * sin 31 – 1.274 * cos 31 This is...

    2 Answers · Science & Mathematics · 06/12/2017

  6. frequency f = 42.0 / 18.9s = 2.22 Hz and ω = 2πf = 14.0 rad/s ω² = k / m, so m = k / ω² = 21.5N/m / (14.0rad/s²) = 0.110 kg ◄ C If you find this helpful, please award Best Answer!

    1 Answers · Science & Mathematics · 30/11/2017

  7. ω = 2π / T √(k/m) = 2π / T k/m = 4π² / T² T² = 4π²m/k...

    1 Answers · Science & Mathematics · 28/11/2017

  8. Actually, this needs to be worked backwards, and you need to FINISH with conservation of momentum. friction work Wf = µmgd = 0.650 * (2600+1500)kg * 9.8m/s² * 6.00m = 156 702 J so this must have been the post-collision KE: 156 702...

    2 Answers · Science & Mathematics · 30/11/2017

  9. ...center to any corner is D = 2d = L*√3/3 Σ V = k*(qA+qB+qC) / (L√3/3) = 3k*(qA+qB+...

    1 Answers · Science & Mathematics · 28/11/2017

  10. ...find the spring compression Kinetic energy = spring energy 1/2*(m1+m2)*v^2 = 1/2*k*x^2 X = v*sprt((m1+m2)/k) = 0.28 m B) initial energy = 1/2*m1*v1i^2...

    2 Answers · Science & Mathematics · 27/11/2017

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