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:
0.375 m/s north & 0.375 m/s east
Explanation:
Answer:
a) -41.1 Joule
b) 108.38 Kelvin
Explanation:
Pressure = P = 290 Pa
Initial volume of gas = V₁ = 0.62 m³
Final volume of gas = V₂ = 0.21 m³
Initial temperature of gas = T₁ = 320 K
Heat loss = Q = -160 J
Work done = PΔV
⇒Work done = 290×(0.21-0.62)
⇒Work done = -118.9 J
a) Change in internal energy = Heat - Work
ΔU = -160 -(-118.9)
⇒ΔU = -41.1 J
∴ Change in internal energy is -41.1 J
b) V₁/V₂ = T₁/T₂
⇒T₂ = T₁V₂/V₁
⇒T₂ = 320×0.21/0.62
⇒T₂ = 108.38 K
∴ Final temperature of the gas is 108.38 Kelvin
Acceleration will be 9.81 if it goes downwards. If it accelerates upwards it will be -9.81m/s^2