Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
Science based on properties of matter and energy
Answer:
Ecu/Eag = 0.46
Explanation:
E = PI/A
Ecu = Pcu × I/A
Pcu = 1.72×10^-8 ohm-meter
r = 0.8 mm = 0.8/1000 = 8×10^-4 m
A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π
Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1
Eag = Pag × I/A
Pag = 1.47×10^-8 ohm-meter
r = 0.5 mm = 0.5/1000 = 5×10^-4 m
A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π
Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π
Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46