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  1. the equation to consider is: q = qo e^- (t/RC) then e^(t/RC) = qo/q t/RC = ln(qo/q) t = RC ln(qo/q) t = 10^3*2x10^-5*ln(20/10) =1.386x10^-2 s

    1 Answers · Science & Mathematics · 25/01/2021

  2. Supposing "standard conditions" to be 1 atm and 0◦ C : (29 mL) x (0.32 atm / 1 atm) x (273 K ) / (27 + 273) K = 8.4 mL

    1 Answers · Science & Mathematics · 23/01/2021

  3. ...should take you.... 3 easy equations to solve this.  if you need more help...  ... . C = 0.034 M / atm * 6.00 atm = X .. .. X * 0.475L = Y...

    1 Answers · Science & Mathematics · 26/01/2021

  4. ... k /m so k = 29.08² * 0.327 = 276.7 Force Constant  k = 277N/m <<<<...0.230m  <<<< ( c )

    1 Answers · Science & Mathematics · 22/01/2021

  5. ...5 C q₃ = -15.1 µ C = -1.51e-5 C The locations are legible. q₃...quot;right" an "up" as positive, F₃₁ = k *q₁*q₃ / d² .... = 8.99e9N·m²...

    1 Answers · Science & Mathematics · 25/01/2021

  6. With a frictionless surface, the ball merely slides (not rolls); the initial spring energy is converted into PE: ½kx² = mgh ½ * 500N/m * (0.25m)² = 1kg * 9.8m/s² * h → → h = 1.6 m

    1 Answers · Science & Mathematics · 22/01/2021

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