The speedA horizontal pipe narrows from a radius of 0.250 m to 0.100 m. If the speed of the water in the pipe is 1.00 m/s in the larger-radius...

We can calculate the average speed with the following relationship:

d^2=4*V/pi*v

d=diameter of the pipe section

V=volume of water

pi=3.14

v= speed of water

Now, we'll write the relation for the section of the pipe which has the radius r1=0.250 m.

d1^2=(2*r1)^2=4*V/pi*v1

4*(0.250)^2=4*V/3.14*1

0.250*3.14/4=V

V=0.196 m^3

We have to know that the same volume of water will crosses the larger section of the pipe, also the smaller narrowed section of the pipe.

So, 4*r2^2*3.14*v2=0.196

4*0.100^2*3.14*v2=0.196

v2=0.196/0.040*3.14

v2=1.56 m/s

Being given the fact that the same volume of water flows through the pipeline, both in the narrow area, also in the wide area, water velocity in the narrowed area will increase.