It's 3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.
Does that tell you anything ?
Why not ?
Answer:
We kindly invite you to read carefully the explanation and check the image attached below.
Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (
,
,
,
)
(2)
Free fall (
,
,
,
)
(3)
Now we created the graph speed-time, which can be seen below.
A path of inferences guided to be cherry picked as for which ones were reasonable and which ones had no ability in the real world to sustain in scientific law
Answer:
Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.
P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0
Explanation:
P1 = P2 = atmospheric pressure so, P1 - P2 = 0
h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²
From the continuity equation for fluids
A1v1 = A2v2
v2 = A1v1/A2
Substituting into the equation above
(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho
Making A2² the subject of the formula,
A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}
The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.
Thank you for reading.