T-Mobile Wi-Fi 9:21 PM 48% . A small, spherical bead of mass 3.10 g is released
ID: 2030876 • Letter: T
Question
T-Mobile Wi-Fi 9:21 PM 48% . A small, spherical bead of mass 3.10 g is released from rest at t-0 from a point under the surface of a viscous liquid. The terminal speed is observed to be VT 1.70 cm/s. (a) Find the value of the constant b in the equation R-b Ns/m (b) Find the time t at which the bead reaches 0.632VT (c) Find the value of the resistive force when the bead reaches terminal speed. Need Help? Determine magnitude of the gravitational force that you exert on another person 2.20 m away. Assume you and the other person each have a mass of 78.5 kg. Need Help? A small piece of Styrofoam packing material is dropped from a height of 1.60 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by ag-Bv. After falling 0.400 m, the Styrofoam effectively reaches terminal speed, and then takes 5.10 s more to reach the ground. (a) What is the value of the constant B? ST (b) What is the acceleration at t-0? m/s2 (down) (c) What is the acceleration when the speed is 0.150 m/s? m/s2 (down) Need Help? .S The system shown in the figure below has an acceleration of magnitude 1.05 m/s2, where m3.80 kg and m2 8.30 kg. Assume that the coefficient of kinetic friction between block and incline is the same for bothExplanation / Answer
7A.
At terminal velocity
R = Vt*b = m*g
b = m*g/Vt
b = 3.1*10^-3*9.81/(1.70*10^-2)
b = 1.79 N-s/m
7B.
relation b/w terminal velocity and time is given by:
V = Vt*(1 - exp(-bt/m))
When V = 0.632*Vt
0.632*Vt = Vt*(1 - exp(-1.79*t/0.0031))
exp(-1.79*t/0.0031) = 1 - 0.632 = 0.368
t = (-0.0031/1.79)*ln 0.368
t = 1.73*10^-3 sec
7C.
At terminal velocity
R = Vt*b
R = 1.70*10^-2*1.79
R = 3.04*10^-2 N
8.
Force is given by:
F = G*m1*m2/R^2
F = 6.67*10^-11*78.5*78.5/2.2^2
F = 8.50*10^-8 N
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