Answer:
a) 0.014N
b) 0.069N
c) 22 seconds
Explanation:
We are given:
t = 49 seconds
[tex] F = 1.4g = 1.4*10^-^3 Kg [/tex]
let's take [tex] g = 9.8m/s^2 [/tex]
a) At terminal speed, net force will be zero.
Therefore,
[tex] F_a_i_r = m * g[/tex]
[tex] F_a_i_r = (1.4*10^-^3kg)(9.8 m/s^2)= 0.014N [/tex]
2. Since 5 coffee filters were dropped, we assume the gravity force and mass will increase by 5, so we have:
[tex] F_a_i_r = 5 * 1.4*10^-^3 *9.8 [/tex]
= 0.069N
3. At terminal velocity, drag force will be a function of V²
Therefore,
[tex]mg=\frac{1}{2}kv^2 [/tex]
Terminal velocity will now be [tex]= \sqrt{m} [/tex]
We are told it takes 49 seconds for 1 filter to hit the ground, thus for 5 filters to hit the grround we have:
[tex] F_d_r_a_g = \frac{49}{\sqrt{5}} [/tex]
= 21.9 => 22 seconds
It will take approximately 22 seconds for a stack of 5 filters to hit the ground
