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:
New volume of the baloon is 0.02325m^3
Explanation:
To answer this question we need to know the ideal gas law, which says:
p•V = n•R•T
p is pressure, V is volume, n is amount of substance (in moles), R is constant value and T is temperature.
Since it's stated that n and T are constant, and we know that R is a constant too, that means that p•V = constant value. Basically, that means that p1•V1 (pressure and volume before the pressure increase) equals to p2•V2 (pressure and volume after the pressure increase).
That means that:
100000 Pa • 0.0279 m^3 = 120000 Pa • V2. Next, V2= 100000•0.0279/120000. So, V2=0.02325m^3.
Time = (distance) / (speed)
Time = (4.12x10^16 m) / (3 x10^8 m/s)
Time = 1.37 x 10^8 seconds
Divide the seconds by 86,400 to get days. Then divide the days by 365 to get years.
Time = about 4.35 years
Answer:
wha is this .....................? same confusion
The mass of an object has no effect whatsoever on the object's
acceleration during free-fall. If there is no air resistance to interfere
with the natural effects of gravity, then a feather and a battleship ...
dropped at the same time ... fall together, and hit the ground at the
same time.
