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1. ### If a 20 kg of mass, m, is attached to the above spring in problem 20 (spring constant k), what will the frequency of oscillation, f, be?

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

2. ### A mass-spring system oscillates with an amplitude of 5.80 cm. If the spring constant is 275 N/m and the mass is 582 g,?

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^2

2 Answers · Science & Mathematics · 08/08/2019

3. ### how far is q2 from q3?

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

4. ### Physics Problem?

l = 10cm = 0.1 m k = 0.84 J/s·m·C A = 30 m² ΔT = 360 ...

2 Answers · Science & Mathematics · 15/08/2019

5. ### a .423 kg object carries a 11.5 charge it is 2.95 M from a -7.55 charge what is the magnitude of the objects acceleration?

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

6. ### Physics question?

... 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

7. ### Physics help??

... it as a constant is very small. So use 4.2 * 10 ^ 3 J/kg/K Energy = dT * m * specific heat = 41 * 55.5 * 4.2 * 10 ^ 3 no of kw...

2 Answers · Science & Mathematics · 03/08/2019

8. ### physics help!A 1.6 μC charge and a 3.1 μC charge are 15.0 cm apart?

...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

9. ### A body that weighs 300 lb. is to be supported on a plane surface which is inclined so that the angle: please see picture - please show work?

The sum of forces F = W (sin(theta) - k cos(theta)) - P = 0 when not sliding. So P = 300*(sin(radians(45)) - .3*cos(radians(45))) = 148.5 lbs. ANS.

2 Answers · Science & Mathematics · 25/07/2019

10. ### find the magnitude of total electric field at point p?

...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