Problem 38 - Elementary Algebra

Question

The velocity \(V\) in feet per second of outflow of a liquid from an orifice is given by \(V=8 \sqrt{h},\) where \(h\) is the height in feet of the liquid above the opening. (a) Determine the velocity of outflow of a liquid from an orifice that is 9 feet below the top surface of a liquid \((V\) is in feet \(/ \mathrm{sec})\). (b) Determine how high a liquid is above an orifice if the velocity of outflow is 81 feet \(/\) second.

Step-by-Step Solution

Answer

Answer: The velocity of outflow of the liquid from an orifice 9 feet below the top surface is 24 feet/sec, and the height of the liquid above the orifice when the velocity of outflow is 81 feet/sec is 6561/64 feet.

Step 1: a) Calculate the velocity of outflow of a liquid from an orifice that is 9 feet below the top surface of a liquid.

Given the height h = 9 feet, we need to calculate the velocity of outflow V. We already have the formula: \(V=8\sqrt{h}\). Substitute h = 9 in the formula: \(V=8\sqrt{9}\). Now, simplify the equation: \(V=8*3 = 24\) feet/sec. Therefore, the velocity of outflow of the liquid from an orifice that is 9 feet below the top surface is 24 feet/sec.

Step 2: b) Determine the height of a liquid above an orifice if the velocity of outflow is 81 feet/sec.

Given the velocity of outflow V = 81 feet/sec, we need to find the height h. We have the formula: \(V=8\sqrt{h}\). Substitute V = 81 in the formula: \(81=8\sqrt{h}\). Now, we need to solve the equation for h. First, divide both sides by 8: \(\frac{81}{8}=\sqrt{h}\). Next, square both sides to eliminate the square root: \((\frac{81}{8})^2=h\). Simplify the equation: \((\frac{6561}{64})=h\). Therefore, the height of the liquid above the orifice is \(\frac{6561}{64}\) feet when the velocity of outflow is 81 feet/sec.

Key Concepts

Understanding Velocity

Velocity is a measure of how fast something moves in a given direction. It's an essential concept in physics and everyday life. When applied to liquids flowing from an opening, velocity is the speed at which the liquid exits the orifice. This is crucial in problems involving outflow because it helps determine how quickly a liquid drains or fills an area.

In mathematical terms, velocity is expressed as a function of distance and time. It tells you how many feet per second or meters per second an object travels. In our problem, velocity (\(V\)) is given in feet per second, which shows how far the liquid moves away from the orifice in one second.

Using Formulas Effectively

A formula is a mathematical equation that expresses a relationship between variables. It's like a recipe that tells you how different quantities are connected.

In this exercise, we rely on the formula \(V = 8 \sqrt{h}\) to determine the velocity of liquid outflow. The formula includes constants and a variable:

• \(V\) is the velocity in feet per second.
• 8 is a constant that modifies how height influences velocity.
• \(\sqrt{h}\) represents the square root of the height \(h\) of the liquid above the orifice.

Plugging different values for \(h\) or \(V\) into the formula lets us solve for the unknowns. For instance, inserting a known height allows calculation of velocity and vice versa. Understanding the role of each component in the formula is key to using it efficiently.

Exploring the Square Root Function

The square root is a fundamental concept in algebra and is often encountered in calculations involving area, volume, and rate equations. In simple terms, the square root of a number is a value that, when multiplied by itself, gives the original number.

In this specific problem, the square root helps us relate the height of the liquid to its velocity. The formula \(V = 8 \sqrt{h}\) indicates that velocity is directly proportional to the square root of height. This means that as the height increases, the effect on velocity is not linear but depends on the squaring function.

For example, with \(h = 9\), \(\sqrt{9}\) becomes 3. Substituting into the formula confirms that when each unit of height increases, it significantly impacts the velocity due to the square root's scaling effect.

Approaches to Problem Solving

Problem-solving is a vital skill in mathematics and beyond, involving the ability to apply learned concepts to new situations to find a solution. It often follows a logical sequence of steps.

Here’s a simple approach to solving the given problem:

• Identify what is known and what needs to be found. Here, you know either the height or the velocity; you need to find the other.
• Assess the applicable formula: \(V = 8 \sqrt{h}\). Understand which variables you have and which you need to solve for.
• Substitute the known values into the formula and perform the necessary algebraic operations. This might involve multiplying, dividing, and applying the square root or squaring.
• Verify the answer by checking whether it makes sense physically in the context of the problem.

By breaking the problem into manageable steps, solving becomes straightforward. With practice, this method becomes part of a larger toolkit for addressing a wide range of mathematical challenges.