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.
It will provide a clear picture of current system functions before any modifications or improvements are made is a benefit if they use the four-model approach.
<u>Option: D</u>
<u>Explanation:</u>
The four model approach is followed by number of analyst which showcase that they construct the physical and logical model of both current and new system. The most important advantage of such approach is it portrays transparent image of ongoing system, before one apply any modification or variation. This is necessary because the flaws which generated earlier in system may affect the SDLC phases and outcome of such process may result into unsatisfied user by paying additional cost. This can be avoided by taking additional steps which make it worth it. The major disadvantage of such approach is the added time and cost of constructing model in both current system.
Answer:
1.89*10⁶J
Explanation:
because 4500kg * 42m*10m/s² is 1890000J
For the answer to the question above, I assume that the question is two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3,?<span>two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3, is electrically neutral.
</span>From Gauss's law:
<span>Flux = ∫c E . dA = q/eo </span>
<span>Since this surface encloses all </span>
<span>charge, we can simplify: </span>
<span>Flux = (q1+q2+q3)/eo </span>
<span>Flux = </span>
<span>( (1*10^-6)+(-1.9*10^-6)+(0) )/(8.85*10^-12) = -101694.92 N·m2/C</span>
