(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
The formation of a cell plate is a characteristic of cytokinesis in terrestrial plants.
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Answer:
The kangaroo was 1.164s in the air before returning to Earth
Explanation:
For this we are going to use the equation of distance for an uniformly accelerated movement, that is:
Where:
x = Final distance
xo = Initial point
Vo = Initial velocity
a = Acceleration
t = time
We have the following values:
x = 1.66m
xo = 0m (the kangaroo starts from the floor)
Vo = 0 m/s (each jump starts from the floor and from a resting position)
a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)
t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.
Now replace the values in the equation
It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is
t = 0.582s + 0.582s
t = 1.164s
The kangaroo was 1.164s in the air before returning to Earth
