3. A 0.400 kg object is released from rest while immersed within a fluid. The object is subject to 2 forces while it des
Posted: Wed May 18, 2022 9:06 am
3. A 0.400 kg object is released from rest while immersed within a fluid. The object is subject to 2 forces while it descends within the fluid: 1. The force due to gravity and 2. A resistive force having a magnitude of by where b is a constant and v is the speed of the mass. a) If the time constant of the system is 0.670 s, what is the constant, b, equal to? (T => 0.6706 0.400 b b=0.400 10.597 kls OCTOS Vy = mg b. What is the terminal speed of the mass? 3109.8) b 0.592 149.25 mis こて
c) At what time after being released will the speed of the mass be 3.50 m/s? Vr = 49.25 ml T= 0.6703 6. 0.597 3150 = 4,925 (1-0 / 0.670s) 3:50= b[VT (1-e -+/0.670)] 3.50 buy t-e-t10070 3.50 2 - t 0:170 e +/0.670 | bur le (1-336 t = -0.670 en 11-18.597) 49.25 10.0855 R max -0.400) (4.6) 3.42N 3.50 d) What is the acceleration of the mass exactly 3.00 s after being released? Vio 14 = 98.01 t 35 Vtry = 49.25(1-e-5/6.070) -48.14 ms Uf = att vi 46.69 = a (3)+8 3 0-16.23 m/? 3
c) At what time after being released will the speed of the mass be 3.50 m/s? Vr = 49.25 ml T= 0.6703 6. 0.597 3150 = 4,925 (1-0 / 0.670s) 3:50= b[VT (1-e -+/0.670)] 3.50 buy t-e-t10070 3.50 2 - t 0:170 e +/0.670 | bur le (1-336 t = -0.670 en 11-18.597) 49.25 10.0855 R max -0.400) (4.6) 3.42N 3.50 d) What is the acceleration of the mass exactly 3.00 s after being released? Vio 14 = 98.01 t 35 Vtry = 49.25(1-e-5/6.070) -48.14 ms Uf = att vi 46.69 = a (3)+8 3 0-16.23 m/? 3