Ideal incompressible water is flowing in a drainage channel of rectangular cross-section. At one point, the width of the channel is 12 m, the depth of water is 6.0 m, and the speed of the flow is 2.5 m/s. At a point downstream, the width has narrowed to 9.0 m, and the depth of water is 8.0 m. What is the speed of the flow at the second point?a. 3.0 m/s b. 2.0 m/s c. 4.0 m/s d. 2.5 m/s e. 3.5 m/s

(D) The speed of the flow at the second point is 2.5 m/s

Explanation:

Given;

at upstream;

the width of the channel, w = 12 m

the depth of water, d = 6.0 m

the speed of the flow, v = 2.5 m/s

at downstream;

the width of the channel, w = 9 m

the depth of water, d = 8.0 m

the speed of the flow, v = ?

Volumetric flow rate for an Ideal incompressible water is given as;

Q = Av

where;

A is the area of the channel of flow

v is velocity of flow

For a constant volumetric flow rate;

A₁v₁ = A₂v₂

A₁ = area of rectangle = L x d = 12 x 6 = 72 m²

A₂ = area of rectangle = L x d = 9 x 8 = 72 m²

(72 x 2.5) = 72v₂

v₂ = 2.5 m/s

Therefore, the speed of the flow at the second point is 2.5 m/s

when the velocity of a mass on a spring is zero, the acceleration is at a maximum .

1 a

2 b

3 b

4 d

5 d

6 d

explanation: