Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
Answer:
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Answer:less
Explanation:thats the answer
Answer:
Explanation:
given,
diameter of merry - go - round = 2.40 m
moment of inertia = I = 356 kg∙m²
speed of the merry- go-round = 1.80 rad/s
mass of child = 25 kg
initial angular momentum of the system
final angular momentum of the system
from conservation of angular momentum
His speed is exactly (350/27) miles per second ... about 46,667 mph. Wotta guy !