The Dependent Variable, is the factor that is measured.
No sorry, wish I could help
Answer:
Explanation:
25 mm diameter
r₁ = 12.5 x 10⁻³ m radius.
cross sectional area = a₁
Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa
a )
velocity of blood v₁ = .6 m /s
Cross sectional area at blockade = 3/4 a₁
Velocity at blockade area = v₂
As liquid is in-compressible
a₁v₁ = a₂v₂
a₁ x .6 m /s = 3/4 a₁ v₂
v₂ = .8m/s
b )
Applying Bernauli's theorem formula
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²
13328 +190.8 = P₂ + 339.2
P₂ = 13179.6 Pa
= 13179 / 13.6 x 10³ x 9.8 m of Hg
P₂ = .09888 m of Hg
98.88 mm of Hg
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.