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... following equation is used to solve this problem. F =

**k*** Q1 * Q2 ÷ d^2 I use 9 * 10^9 for**k**. The unit of the...2 Answers · Science & Mathematics · 16/08/2019

You should post your answers so (if they are wrong) we can explain why they are wrong. 1a). Mass (m) stays the same. Volume (V) is halved. Since density = m/V, the density doubles. 1b) I don'...

1 Answers · Science & Mathematics · 18/08/2019

f = 1/(2 * π) * √(

**k**/m) The first step is solve problem 20 to determine the spring constant... to determine the increase of the length. ∆ L = 0.15 – 0.10 = 0.05 meter**k**= 4 ÷ 0.05 = 80 N/m Now we can use the spring constant to calculate the...2 Answers · Science & Mathematics · 16/08/2019

F[1,2] = F[2,3] -

**k*** Q[1] * Q[2] / r[1,2]^2 = -**k*** Q[2] * Q[3] / r[2,3...2 Answers · Science & Mathematics · 11/08/2019

1) E = 1/2

**k**A^2 E = 0.5 * 275 * 0.058^2 = 0.46 J 2) v = w A = sqrt(**k**/m) A v = sqrt(275/0.582) * 0.058 = 1.26 m/s 3) a = w^2 A = (**k**/m) A a = (275/0.582) * 0.058 = 27.41 m/s^22 Answers · Science & Mathematics · 08/08/2019

...data are given to 3 sig. figs, we should use the value of ‘

**k**’ to 3 sig. figs. E1 = kq1/d² = 8.99*10^9 * 6.60*10^-9...2 Answers · Science & Mathematics · 08/08/2019

... is force in newtons r is separation in meters

**k**= 8.99e9 Nm²/C²1 Answers · Science & Mathematics · 08/08/2019

EDIT. The force on each charge is: F =

**k**.q1.q2/r² = 9x10^9 x 11.5x10^-6 x (-7.55)x10^-6 / 2.95² = -0.0898...3 Answers · Science & Mathematics · 07/08/2019

a)what is the total angular momentum of the system before the collision? as both rod and dis

**k**have identical mass m2, let's just use the symbol (m) to avoid confusion. assume the center of the length of the rod lands and stays on the center of rotation of the dis**k**. L = ΣIω...1 Answers · Science & Mathematics · 07/08/2019

...between them and cancel where they have equal magnitude. We can factor out

**k**and 1e-6 and write: 1.6/r² = 3.1/(0.15-r)² 1.6(0.15-r)²...3 Answers · Science & Mathematics · 07/08/2019

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