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91Ó°ÊÓ

Niagara Falls has a height of \(167 \mathrm{ft}\) (American Falls). What is the potential energy in joules of \(1.00 \mathrm{lb}\) of water at the top of the falls if we take water at the bottom to have a potential energy of zero? What would be the speed of this water at the bottom of the falls if we neglect friction during the descent of the water?

Short Answer

Expert verified
Potential energy is 226.26 joules; speed at bottom is 30.76 m/s.

Step by step solution

01

Conversion of Units for Mass

Since potential energy is usually calculated with biological units such as kilograms, we need to convert the mass of water from pounds to kilograms. We know that 1 lb is approximately equal to 0.453592 kg. Thus, the mass in kilograms is given by: \(1 \, \text{lb} \times 0.453592 \, \text{kg/lb} = 0.453592 \, \text{kg}.\)
02

Conversion of Units for Height

We need to convert the height from feet to meters because the standard unit in calculations involving joules is the meter. 1 foot is approximately 0.3048 meters. Therefore, the height in meters is: \(167 \, \text{feet} \times 0.3048 \, \text{meters/foot} \approx 50.902816 \, \text{meters}.\)
03

Calculation of Potential Energy

The potential energy \( (PE) \) at the top can be calculated using the formula: \(PE = mgh\), where \(m = 0.453592 \, \text{kg}\), \(g = 9.81 \, \text{m/s}^2\), and \(h \approx 50.902816 \, \text{meters}.\) So, the potential energy is: \(PE = 0.453592 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 50.902816 \, \text{m} \approx 226.26 \, \text{joules}.\)
04

Using Conservation of Energy to Find Speed

The potential energy at the top will be converted into kinetic energy at the bottom, meaning \(PE_{top} = KE_{bottom}\). The kinetic energy \((KE)\) is given by \(KE = \frac{1}{2} mv^2\). So, \(226.26 = \frac{1}{2} \times 0.453592 \times v^2\). Solve for \(v\):\[v^2 = \left(\frac{2 \times 226.26}{0.453592}\right)\]\[v = \sqrt{\frac{2 \times 226.26}{0.453592}} \approx 30.76 \, \text{m/s}.\] The speed of water at the bottom is approximately \(30.76 \, \text{m/s}.\)

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Key Concepts

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

Unit Conversion
Understanding how to convert units is essential in physics, as calculations often require standard units for accuracy. Unit conversion ensures that mathematical equations and principles remain consistent.

**Mass Conversion**
Many physics problems, like the one involving Niagara Falls, require converting mass from pounds to kilograms since the metric system is preferred in scientific calculations.
  • 1 pound (lb) = 0.453592 kilograms (kg)
This means, to convert from pounds to kilograms, multiply the mass in pounds by 0.453592.

**Height Conversion**
Similarly, converting height measurements from feet to meters is crucial as most physics equations, including energy calculations, use meters.
  • 1 foot = 0.3048 meters
By using these conversion factors, you ensure all your measurements align with scientific norms.
Gravitational Potential Energy
Gravitational potential energy is a type of potential energy that depends on an object's position relative to the Earth. It is the energy stored due to the object's height and gravitational force acting on it.

The formula to calculate gravitational potential energy is:
  • \( PE = mgh \), where:
  • \( m \) is the mass in kilograms (kg)
  • \( g \) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\)
  • \( h \) is the height in meters (m)
This formula shows that potential energy increases with both mass and height.

In the context of Niagara Falls, the energy potential was calculated using the converted mass and height to understand how much energy a pound of water possesses at the fall's crest.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. When an object is in motion, it has the potential to do work because of its velocity.

The formula to calculate kinetic energy is:
  • \( KE = \frac{1}{2} mv^2 \), where:
  • \( m \) is the mass in kilograms (kg)
  • \( v \) is the velocity in meters per second (m/s)
This equation shows that kinetic energy increases with the square of velocity, meaning small changes in speed have a large effect.

In the Niagara Falls exercise, it demonstrates how the potential energy of water at the top converts into kinetic energy, giving us the velocity at the waterfall's bottom if friction is ignored.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another.

In the context of the waterfall:
  • The potential energy at the top converts entirely into kinetic energy at the bottom (if we disregard friction).
This means, if you know how much potential energy is there at the start, you could determine the kinetic energy value at the waterfall's base.

Practically speaking, during energy conversion, some energy is often lost to friction or sound. However, assuming ideal conditions helps solve problems more clearly.
Physics Problem Solving
Solving physics problems typically involves a systematic approach to find solutions using known principles. Key steps include:
  • Identifying known and unknown variables in the problem.
  • Converting units where necessary to match standard formulas.
  • Applying relevant formulas to calculate needed values.
  • Considering principles like the conservation laws to relate different forms of energy or forces.
  • Checking the solution's consistency with anticipated results.
In solving the Niagara Falls problem, each step targets a specific task. First, we convert all units to SI units. Next, the potential energy is determined using known values for mass and gravity, then translated into kinetic energy to find speed. This methodical approach ensures comprehensive understanding and accurate results.

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Most popular questions from this chapter

Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, \(\mathrm{NO}\). $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What is the heat of reaction at constant pressure? Use the following thermochemical equations: $$ \begin{gathered} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) ; \Delta H=180.6 \mathrm{~kJ} \\ \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) ; \Delta H=-91.8 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-483.7 \mathrm{~kJ} \end{gathered} $$

Nitrous oxide, \(\mathrm{N}_{2} \mathrm{O}\), has been used as a dental anesthetic. The average speed of an \(\mathrm{N}_{2} \mathrm{O}\) molecule at \(25^{\circ} \mathrm{C}\) is \(379 \mathrm{~m} / \mathrm{s}\). Calculate the kinetic energy (in joules) of an \(\mathrm{N}_{2} \mathrm{O}\) molecule traveling at this speed.

The sugar arabinose, \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\), is burned completely in oxygen in a calorimeter. $$ \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}(s)+5 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{CO}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O}(l) $$ Burning a \(0.548-\mathrm{g}\) sample caused the temperature to rise from \(20.00^{\circ} \mathrm{C}\) to \(20.54^{\circ} \mathrm{C}\). The heat capacity of the calorimeter and its contents is \(15.8 \mathrm{~kJ} /{ }^{\circ} \mathrm{C} .\) Calculate \(\Delta H\) for the combustion reaction per mole of arabinose.

Part 1: In an insulated container, you mix 200. g of water at \(80^{\circ} \mathrm{C}\) with \(100 . \mathrm{g}\) of water at \(20^{\circ} \mathrm{C}\). After mixing, the temperature of the water is \(60^{\circ} \mathrm{C}\). a. How much did the temperature of the hot water change? How much did the temperature of the cold water change? Compare the magnitudes (positive values) of these changes. b. During the mixing, how did the heat transfer occur: from hot water to cold, or from cold water to hot? C. What quantity of heat was transferred from one sample to the other? d. How does the quantity of heat transferred to or from the hot-water sample compare with the quantity of heat transferred to or from the cold-water sample? e. Knowing these relative quantities of heat, why is the temperature change of the cold water greater than the magnitude of the temperature change of the hot water. f. A sample of hot water is mixed with a sample of cold water that has twice its mass. Predict the temperature change of each of the samples. g. You mix two samples of water, and one increases by \(20^{\circ} \mathrm{C}\), while the other drops by \(60^{\circ} \mathrm{C}\). Which of the samples has less mass? How do the masses of the two water samples compare? h. A 7 -g sample of hot water is mixed with a \(3-\mathrm{g}\) sample of cold water. How do the temperature changes of the two water samples compare? Part \(2:\) A sample of water is heated from \(10^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). Can you calculate the amount of heat added to the water sample that caused this temperature change? If not, what information do you need to perform this calculation? Part 3: Two samples of water are heated from \(20^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\). One of the samples requires twice as much heat to bring about this temperature change as the other. How do the masses of the two water samples compare? Explain your reasoning.

Ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\), is used as antifreeze. It is produced from ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), by the reaction $$ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l) $$ Use Hess's law to obtain the enthalpy change for this reaction from the following enthalpy changes: $$ \begin{gathered} 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-2612.2 \mathrm{~kJ} \\ \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) ; \\ \Delta H=-1189.8 \mathrm{~kJ} \end{gathered} $$

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