Let’s look at the falling mouse example we touched on in class in more detail. L
ID: 1592313 • Letter: L
Question
Let’s look at the falling mouse example we touched on in class in more detail. Let’s make things as simple as possible: consider a spherical mouse of radius 0.02 m and a spherical human of radius 0.25 m. These unfortunate creatures have the same density as water, 1000 kg/m3.
(a) The volume of a sphere is Vsphere = 4 3r3. What are the masses of the human and the mouse?
(b) What are the weights of the human and the mouse?
(c) In this particular case, the approximate force of air resistance is given by the following formula:
Fdrag =1/8 ×Density of air×(Speed)^2 ×Surface Area of Sphere
The density of air is 1.225 kg/m3, and the surface area of a sphere is given by the formula SAsphere = 4r^2. At terminal speed, the forces on each animal are balanced—that is, W = Fdrag. Using all of this, nd the terminal speed of both the mouse and the human
Explanation / Answer
Here,
radius of mouse , r1 = 0.02 m
radius of human , r2 = 0.25 m
density , p = 1000 Kg/m^3
a)
mass of mouse = density * volume of mouse
mass of mouse = 1000 * 4/3 *pi * 0.02^3
mass of mouse = 0.0335 Kg
mass of human= density * volume of mouse
mass of human = 1000 * 4/3 * pi * 0.25^3
mass of human =65.45 Kg
b)
weight of mouse = mass * g
weight of mouse = 0.0335 * 9.8
weight of mouse = 0.3283 N
weight of human = 65.45 * 9.8
weight of human = 641.41 N
c)
for mouse
as Fdrag = 1/8 * density of air * speed^2 * area
for weight = drag force
1/8 * 1.225 * speed^2 * 4 *pi * 0.02^2 = 0.3283
solving
speed = 20.65 m/s
the terminal speed is 20.65 m/s
for human
as Fdrag = 1/8 * density of air * speed^2 * area
for weight = drag force
1/8 * 1.225 * speed^2 * 4 *pi * 0.25^2 = 641.41
solving
speed = 70.03 m/s
the terminal speed of human is 70.03 m/s
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