This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — here is the solution to question 4. 4. The figure below shows how water is drawn from a large tank into a low-lying container using a rubber tube of uniform cross-sectional area 0.002 m^2. Determine the force driving out the water. Step 1: Identify the effective height difference. The force driving the water out is due to the pressure difference created by the vertical height difference between the water surface in the large tank and the outlet of the tube. From the diagram: Height of water surface from the top of the tube = 50 cm Height of the outlet from the top of the tube = 150 cm Effective height difference h = 150 cm - 50 cm = 100 cm. Step 2: Convert the height to meters. h = 100 cm = 1.0 m Step 3: State the given values. Density of water = 1000 kg/m^3 Gravitational field strength g = 9.8 N/kg (standard value) Cross-sectional area A = 0.002 m^2 Step 4: Calculate the pressure difference. The pressure difference P is given by the formula P = g h: P = (1000 kg/m^3) × (9.8 N/kg) × (1.0 m) P = 9800 Pa Step 5: Calculate the force driving out the water. The force F is the pressure difference multiplied by the cross-sectional area: F = P × A F = (9800 Pa) × (0.002 m^2) F = 19.6 N The force driving out the water is 19.6 N. Got more? Send 'em.