2.
a)
Let spring K1 be system A and spring K2 with damper be system B.
From spring properties, we know that
F=FA=FB
and
L=L1+L2
Substituting eq(1) and eq(2)
Rearranging like items,
Multiplying both sides by K1,
b)
When length is first stretched, K1takes a
1.
Known:
Where D is the distensibility,
L is the length of the chamber.
a)
Substituting D:
Substituting Vd:
Assuming d=dsys-ddias=ddias,
If not, the approximation would be false.
b)
a)Known:
P=120-80=40mmHg=40*1333dyne/cm2
Ddias=4.0cm
Clinical group
Dist
4.
a)
Since the whole body is in static equilibrium, we know that =0, and using point
O as the center (because the two other unknown F n and Fs both pass this point):
Fe*a-2Wb/3*b-P*c=0
We know that 2Wb/3=660(N)
P=150(N)
a=3cm, b=9cm, c=27cm
Fe*(3)-(660)*
3.
a)
Since the whole leg is in static equilibrium, we know that =0, and using point O
as the center:
Fp*dpo+Fj*doo+Wl*dol+Rg*dgj=0
We know that Wl=120*.07=8.4(kg)
Rg=120/2=60(kg)
Fp*(7.6cm)+Fj*(0cm)+(8.4kg)*(21cm)-(60kg)*(43cm)=0
Solving for Fp=316.26(kg
5.
a)
b) Since the tympanic membrane is connected to the oval window with a rigid rod, they act
together as a body; therefore, when the tympanic membrane is displaced 1m, the oval window
is displaced 1m, and when the tympanic membrane moves with velocity