(a) Estimate the terminal speed of a wooden sphere (density 0.810 g/cm2) falling
ID: 1875886 • Letter: #
Question
(a) Estimate the terminal speed of a wooden sphere (density 0.810 g/cm2) falling through air, if its radius is 7.50 cm and its drag kg/m3.) 25.8453 x Your response differs from the correct answer by more than 10%. Double check your calculations. m/s is 0.500. (The density of air is 1.20 (b) From what height would a freely falling object reach this speed in the absence of air resistance? 34.08071 Your response differs from the correct answer by more than 10%. Double check your calculations. m Need Help?Explanation / Answer
|Fair | = cAv2, where c is a constant, is the density of the air, A is the cross-sectional area of the object as seen from below, and v is the object's velocity. Predict the object's terminal velocity, i.e., the final velocity it reaches after a long time.
Fair + F(downwards) = 0 , i.e., --------------------------1)
1/2*cAv2 - mg = 0 gravitational force acting on wooden sphere will be counter acted by Fair to to reach terminal velocity using eqn 1)
Now as we know (mass of wooden block = Volume of sphere x Density)
m = 4/3*pi*(7.5)^3 x 0.81 ; V = 4/3*pi*r^3 , where r = 7.5 cm, pi = 22/7
m = 1431.96 g = 1.43196 kg
and cross sectional area of wooden block A = pi*r^2 = 0.0176 m^2
now from eqn
(1/2)cAv2 - mg = 0 terminal vel "v" = ?, A = 0.0176 m^2, c = 0.5, = 1.2 kg/m^3
0.5*1.2*0.0176*v^2 = 2*1.43196 * 9.81 ; (g = 9.81 m/sec^2)
v^2 = 2660.51
v(terminal) = 51.580 or 51.580 m/sec or --------------- Answer
b) Suppose there is no resistance and object is falling strictly under gravitational force
we have v^2 = u^2 + 2gh (3rd law of motion, where v = final vel = v(terminal) , u = 0, h = ?
(51.580)^2 = 0 + 2*9.81*h
h = 135.60 m -------------
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.