Water is flowing in a pipe with a varying cross-sectional area, and at all points, the water completely fills the pipe. at point 1, the cross-sectional area of the pipe is 7.70×10−2 m2 and the magnitude of the fluid velocity is 3.60 m/s.
a. what is the fluid speed at points in the pipe where the cross-sectional area is (a) 0.105 m and (b) 0.047m2?
b. calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

a. 2.64 m/s

b. 5.9 m/s

c. 997.92 m3

Explanation:

As this is steady flow, the mass flow rate is constant, and so is the product of flow velocity and cross-sectional area

av = 0.077 * 3.6 = 0.2772 m3/s

We can calculate the speed at various areas by dividing the product above by area

a.

b.

c. Since the volume discharge rate is 0.2772 cube meters per second. In 1 hour, or 60 * 60 = 3600 seconds, the total volume of water discharged would be

0.2772 * 3600 = 997.92 m3

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation:

so easy. just google the electromagnetic waves and it’ll pop up.