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:
Explanation:
<u>Vertical Launch Upwards</u>
In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:
The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.
(a)
To find the initial speed we solve for vo:
(b)
The maximum time or the time taken by the object to reach its highest point is calculated as follows:
Answer:
A maximum
Explanation:
When displacement is maximum, velocity is Zero and vice versa
When displacement is maximum, acceleration is maximum and when it is zero, acc. Is zero
If the bubble travels 10 meters per second and it takes 10 seconds, then just multiply the distance per second by the total seconds to get the total depth.
10 • 10 = 100
The lake is 100 meters deep.
Think of it this way to clarify the answer:
It takes a bubble traveling at a speed of 10 meters per second 10 seconds to travel 100 meters.
Answer:
y = 128.0 km
Explanation:
The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains
the diffraction equation for the first minimum is
a sin θ = λ
In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.
θ = 1.22 λ / d
d = 15 cm
to find the distance we can use trigonometry
tan θ = y / L
tan θ = sin θ / cos θ = θ
substituting
y / L = λ / d
y = L λ /d
let's calculate
y = 384 10⁸ 500 10⁻⁹ / 0.15
y = 1.28 10⁵ m
Let's reduce to km
y = 1.28 10⁵ m (1km / 10³ m)
y = 128.0 km
the correct answer is 120 km away
