Astiff, 15-cm-long tube with an inner diameter of 2.5 mm is attached to a small hole in the side of a tall beaker. the tube sticks out horizontally. the beaker is filled with 20∘c water to a level 45 cm above the hole, and it is continually topped off to it maintain that level. what is the volume flow rate through the tube?

Volume rate of flow through the pipe is 1.457 × 10⁻⁵ m³/s

Explanation:

Given:

Inner diameter of the tube, d = 2.5 mm = 2.5 × 10⁻³ m

water level above the hole, h = 45 cm = 0.45 m

now,

the velocity of the water at the level of hole, v = √(2gh)

thus,

v = √(2 × 9.8 ×.45)

or

v = 2.969 m/s

Now,

the discharge (Q) is given as:

Q = Area × velocity

area of the outlet  = πd²/4

or

area = π(2.5 × 10⁻³)²/4 = 4.90 × 10⁻⁶ m²

on substituting the values in the formula for discharge, we get

Q = 4.90 × 10⁻⁶ m² × 2.969 m/s

or

Q = 1.457 × 10⁻⁵ m³/s

hence, the volume rate of flow through the pipe is 1.457 × 10⁻⁵ m³/s

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.