Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
Answer:
Explanation:
a ) The volume of blood flowing per second throughout the vessel is constant .
a₁ v₁ = a₂ v₂
a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .
2A x v₁ = A x .40
v₁ = .20 m /s
b )
Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂
Applying Bernoulli's theorem
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
P₁ - P₂ = 1/2 ρ(v₂² - v₁² )
= .5 x 1060 ( .4² - .2² )
= 63.6 Pa .
Weight = 45 * 9.8 = 441 N
I think it’s C but not sure..
