Apec Gaming asked in Science & MathematicsPhysics · 4 months ago

Physics Help Plx (plz provide answer and not formula) THX!!!!?

A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 4.0m/s.

What is the speed in the second segment?

What is the speed in the third segment?

What is the volume flow rate through the pipe?

Relevance
• Vaman
Lv 7
4 months ago

we must use r^2 v= constant. r is the radius and v is the velocity.

first condition gives 1/4*4=1. This remains a constant. In the second segment r=1. velocity =1 m/s. In the third segment (1/4)^2 v=1.

v=16 m/s. In the first segment it is 4 m/s, in the second it is 1 m/s, in the third it is 16 m/s.

• JOHN
Lv 7
4 months ago

Every second the same quantity of fluid must pass through every cross-section of the pipe.

Let u =4.0m/s, v, w be the fluid velocities in the 1cm, 2cm and 0.5cm sections of the pipe.

In 1s π(1/2)² x 400cm³ passes any 1cm diameter cross-section. For the 2cm section this is π(2/2)²v = π(1/2)² x 400, giving v = 100cm/s = 1m/s.

For the 0.5cm section this is π(0.5/2)²w = π(1/2)² x 400, giving w = 1600cm/s = 16m/s.