Answer
given,
mass of the drop, m = 0.0014 g
speed of the drop, u = 8.1 m/s
a) Change in momentum is equal to impulse
final velocity of the drop, v = 0 m/s
J = m ( v - u )
J = 0.0014 x 10⁻³ x ( 0 - 8.1 )
J = -1.134 x 10⁻⁵ kg.m/s
impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s
b) time, t = 0.37 m s
impact of force = ?
we know
J = F x t
1.134 x 10⁻⁵ = F x 0.37 x 10⁻³
F = 0.031 N
the magnitude of the force of the impact is equal to F = 0.031 N
Answer:
m' = 1 x 10⁻⁶ kg/s
Explanation:
Given that
Diffussion constant = 1 x 10⁻¹¹
Mass flow rate ,m = 2 x 10⁻⁶ kg/s
The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.
So new flow rate m'


m' = 1 x 10⁻⁶ kg/s
Answer:
net force would be 50 N right
Explanation: