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

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

(8 atm) x (333 + 273)

**K**/ (83 + 273)**K**= 13.618 = 14 atm2 Answers · Science & Mathematics · 23/01/2021

...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

... 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 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

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