1) The terminal velocity of a 4×10^−5 kg raindrop is about 5 m/s . Assuming a drag force Fd= -bv.
A) Assuming a drag force determine the value of the constant b.
B) Assuming a drag force determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.
PLEASE EXPLAIN. 10 POINTS!!
Dr. Watkin,. I presume
Favorite Answer
A)
At the terminal velocity the downward force of gravity balances the upward drag force, so the net force on the raindrop is 0 and it falls at constant speed.
So Fd = -m*g = -4*10^-5kg * 9.8 m/s^2 = -3.92*10^-4 N
Also Fd = -b*v when v = 5 m/s
so b = -Fd / v = 3.92*10^-4 / 5 = 7.84*10^-5 N.s/m <= ANS
B)
The force of gravity is reduced by the drag force, so the free-fall acceleration from rest will be
a = g - Fd/m instead of g.
v = 63% of 5m/s = 3.15 m/s
v = a*t = (g - Fd/m)*t = 3.15 m/s
At v=3.15 m/s Fd = -b*v = 2.47*10^-4 N
so Fd/m = 6.17 m/s^2 (= reduction in acceleration)
Hence t = v / a = 3.15 / (9.8 - 6.17) s = 0.87 s <= ANS
?
The arrow keys ought to administration course - that's the organic and organic inclination. Use J for leaping, F to increasw stress, shift F or ctrl-F to cut back it, A for acceleration, V for velocity (assuming which you at the instant are not portraying rather physics - indoors the rather international, increasing stress creates acceleration that will boost velocity).
Anonymous
All physics is drag force physics...it's a drag, do the math and figure out and what force, lol