Answer:
The average speed of the blood in the capillaries is 0.047 cm/s.
Explanation:
Given;
radius of the aorta, r₁ = 1 cm
speed of blood, v₁ = 30 cm/s
Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²
Area of the capillaries, A₂ = 2000 cm²
let the average speed of the blood in the capillaries = v₂
Apply continuity equation to determine the average speed of the blood in the capillaries.
A₁v₁ = A₂v₂
v₂ = (A₁v₁) / (A₂)
v₂ = (3.142 x 30) / (2000)
v₂ = 0.047 cm/s
Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.
Answer:
1.05m or 105cm
Explanation:
Using the hooke's law equation as follows;
F = –k.x
Where;
F = force (N)
x = extension length (m)
k = constant of proportionality (N/m)
According to the information given in this question;
Displacement (x) = 85cm = 85/100 = 0.85m
Force = 12500N
Using F = kx, we find the proportionality constant
k = F/x
K = 12500/0.85
K = 14705.8N/m.
Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;
F = kx
x = F/k
x = 15500/14705.8
x = 1.05m or 105cm
Answer:
The diameter of the camera aperture must be greater than or equal to 1.49m
Explanation:
Let the distance separating two objects, x = 6.0 cm = 0.06 m
The distance between the observer and the two objects, d = 160 km = 160000 m
Let ∅ = minimum angular separation between the two objects that the satellite can resolve
tan( ∅) = x/d
Since there is minimum angular separation, tan( ∅) ≈∅
∅ = x/d
∅ = 0.06/160000
∅ = 3.75 * 10⁻⁷rad
For the satellite to be able to resolve the objects,
D ≥ 1.22λ/∅
λ = 560 nm = 560 * 10⁻⁹
D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)
D ≥ 149.33 * 10⁻² m
D ≥ 1.49 m
