Answer:
Explanation:
mass of coffee filter, m = 1.4 h
height, h = 4 m
velocity, v = 0.5 m/s
The energy conservation
Potential energy of filter = kinetic energy of filter + energy gained by the air molecules
m x g x h = 0.5 x mv² + energy gained by the air molecules
0.0014 x 9.8 x 4 = 0.5 x 0.0014 x 0.5 x 0.5 + Energy gained by the air molecules
Energy gained by the air molecules = 0.05488 - 0.00000175
= 0.054878 J
Answer:
0.055 J
Explanation:
We are given that
Mass=m=1.4 g=[tex]1.4\times 10^{-3} kg[/tex]
1 kg=1000 g
h=4 m
v=0.5 m/s
Potential energy=[tex]mgh[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Using the formula
Potential energy of coffee filter lost=[tex]1.4\times 10^{-4}\times 9.8\times 4=5.5\times 10^{-2} J[/tex]
Kinetic energy of coffee filter gain=[tex]\frac{1}{2}mv^2=\frac{1}{2}(1.4\times 10^{-3})\times (0.5)^2=1.75\cdot 10^{-4} J[/tex]
Kinetic energy gain by air molecules from the falling coffee filter=Potential energy-Kinetic energy
Kinetic energy gain by air molecules from the falling coffee filter=[tex]5.5\times 10^{-2}-1.75\times 10^{-4}=0.055 J[/tex]
Hence, the kinetic energy gain by air molecules from the falling coffee filter=0.055 J
