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  1. 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. ...2 ] + 0.5k[y(t) - x(t) - a]^2 = 0.5m(v_0)^2 (v_0)^2 - (dx/dt)^2 - 2(dy/dt)^2 = (k/m)[y(t) - x(t) - a]^2 4. When dx/dt = 0 then dy/dt = v_0 / 2 Plug...

    1 Answers · Science & Mathematics · 28/06/2019

  3. For diatomic ideal gas U = 5/2 nRT (should be Delta on U and T) U = Q - W T = (Q-W)/(5/2 nR) = (390-185)/(5/2 0.21*3.814)

    1 Answers · Science & Mathematics · 23/05/2019

  4. Let’s use the following equation to determine the final pressure of the gas. P1 ÷ T1 = P2 ÷ T2 2.37 ÷ 273 = P2 ÷ 610 P2 * 273 = 1,445.7 P2 =1,445.7 ÷ 273 This is approximately 5.3 atm. I hope this is helpful for you.

    2 Answers · Science & Mathematics · 12/05/2019

  5. ... energy is conserved PE max = PE at 2cm + KE at 2cm 1/2(1400)(0.0500^2) = 1/2(1400)(0.0200^2) + 1/2 (1.50) v^2 1/2 's cancel 3.5 = 0.56 + 1.50 v^2 1.50 ^2 = 2.94 v^2 = 1.96 v = 1.4 m/s When you get a good response, please consider giving a best answer. This...

    1 Answers · Science & Mathematics · 17/04/2019

  6. two 4.4 in series are equal to 2.2 two 2.5k in series are equal to 5 time constant = 2.2µ x 5k = 11 ms voltage across 5k is 1.5•5 = 7.5 volts voltage across cap = 11.4–7.5 = 3.9 volts v = v₀[1–e^(–t/τ)] 3.9 = 11.4[1–e^(–t/11m)] 1–e^(–t/11m) = 0.3421 e^(–t/11m) = 0.6579 –t/11m...

    1 Answers · Science & Mathematics · 02/05/2019

  7. Assuming the system has no friction, it's a simple harmonic oscillator and therefore has constant mechanical energy. When the spring is fully stretched, all of that mechanical energy is spring potential E = PS E = ½kA²...

    2 Answers · Science & Mathematics · 28/05/2019

  8. At equilibrium position, there is no spring potential energy. E = 0.5mv^2 = 0.5*0.15*2.39^2 = 0.428 J

    2 Answers · Science & Mathematics · 28/05/2019

  9. During the time that the pebble is being launched, there are two forces on the pebble. The force exerted by the toy gun is the upward force. Its weight is the downward force. Let’s determine the work that is done...

    2 Answers · Science & Mathematics · 28/01/2019

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