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A stream of water flowing horizontally with a speed of \(15 \mathrm{~m} \mathrm{~s}^{-1}\) gushes out of a tube of cross-sectional area \(10^{-2} \mathrm{~m}^{2}\), and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Short Answer

Expert verified
The force exerted on the wall is 2250 N.

Step by step solution

01

Identify the relevant physics principle

The problem involves water impacting a wall, which relates to momentum and force concepts. The force exerted by the water can be found using the rate of change of momentum.
02

Calculate the volume flow rate

Find the volume of water hitting the wall per second by multiplying the cross-sectional area of the tube by the velocity of the water. \[\text{Volume flow rate} = A \times v = 10^{-2} \, \text{m}^2 \times 15 \, \text{m/s} = 0.15 \, \text{m}^3/\text{s}\]
03

Calculate the mass flow rate

The mass of water hitting the wall per second is the volume flow rate multiplied by the density of water (\(1000 \, \text{kg/m}^3\)). \[\text{Mass flow rate} = \text{Volume flow rate} \times \text{density} = 0.15 \, \text{m}^3/\text{s} \times 1000 \, \text{kg/m}^3 = 150 \, \text{kg/s}\]
04

Determine the force exerted on the wall

The force exerted on the wall is given by the change in momentum per second. Since the water stops when it hits the wall, the final velocity is zero. Therefore, the change in momentum per second is:\[F = \text{Mass flow rate} \times \text{change in velocity} = 150 \, \text{kg/s} \times 15 \, \text{m/s} = 2250 \, \text{N}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force
Force is a fundamental concept in physics that describes the interaction between objects. It's a push or pull action that can cause an object to accelerate, decelerate, remain in equilibrium, or change its state of motion. In simple terms, force is what makes things move or stop moving.
Force is mathematically defined as the product of mass and acceleration, given by Newton’s second law of motion:
  • \( F = m \times a \)
Here, \( F \) is the force in newtons (N), \( m \) is the mass in kilograms (kg), and \( a \) is the acceleration in meters per second squared (\( m/s^2 \)).
In the context of the problem, force is calculated using the change in momentum over time, as the water hits the wall. Once the water stops, the change in velocity is the negative of the initial velocity (assuming the water doesn't bounce back). This change is used in conjunction with the mass flow rate to determine the force exerted on the wall.
Mass Flow Rate
Mass flow rate is a measure of the amount of mass passing through a given surface per unit time. It’s typically used in fluid dynamics to understand how quickly a fluid is moving through a pipe or opening. It is a crucial parameter in systems where mass is transmitted continuously, like in the exercise given.
The mass flow rate \( \dot{m} \) can be expressed as:
  • \( \dot{m} = \rho \times Q \)
Where \( \rho \) is the fluid density (in kilograms per cubic meter, \( \text{kg/m}^3 \)) and \( Q \) is the volume flow rate (in cubic meters per second, \( \text{m}^3/s \)).
For water with a density of \( 1000 \text{kg/m}^3 \), the mass flow rate becomes straightforward to calculate. It helps quantify how much mass is impacting the wall every second, which is then used to determine the force exerted on the wall, by knowing the velocity change upon impact.
Volume Flow Rate
The volume flow rate is used to measure how much fluid, in terms of volume, passes through a point in a system per unit of time. Especially in fluid dynamics and engineering, it's significant for systems involving pipelines or channels.
Mathematically, volume flow rate \( Q \) is defined as:
  • \( Q = A \times v \)
Here, \( A \) is the cross-sectional area through which the fluid flows (in square meters, \( \text{m}^2 \)) and \( v \) is the velocity of the fluid (in meters per second, \( \text{m/s} \)).
In scenarios like the exercise given, the volume flow rate tells us exactly how much water is striking the wall each second. This value becomes critical in calculating the mass flow rate and eventually the force. It especially helps in visualizing the constant flow in terms of how much space the fluid would occupy as it moves.

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