What is the reactive force when a fish swims through water?
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By vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
where
is its direction with respect to the x-axis.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
where
To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.
The work done would be defined as
Where,
PE = Potential Energy
KE = Kinetic Energy
Where,
m = Mass
g = Gravitational energy
h = Height
v = Velocity
Considering power as the change of energy as a function of time we will then have to
The rate of mass flow is,
Where,
= Density of water
A = Area of the hose
The given radius is 0.83cm or m, so the Area would be
We have then that,
Final the power of the pump would be,
Therefore the power of the pump is 57.11W