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At sea level, a nitrogen molecule in the air has an average kinetic energy of \(6.2 \cdot 10^{-21}\) J. Its mass is \(4.7 \cdot 10^{-26} \mathrm{~kg}\). If the molecule could shoot straight up without colliding with other molecules, how high would it rise? What percentage of the Earth's radius is this height? What is the molecule's initial speed? (Assume that you can use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\); although we'll see in Chapter 12 that this assumption may not be justified for this situation.

Short Answer

Expert verified
Also, what is its initial speed? Answer: The nitrogen molecule would rise approximately 1.346 x 10^5 m high, representing about 0.02114% of the Earth's radius. Its initial speed is approximately 391.62 m/s.

Step by step solution

01

Find the initial velocity of the molecule

We can find the initial velocity (v) using the kinetic energy formula: \(KE = \frac{1}{2} m \cdot v^2\) Where KE is the kinetic energy, m is the mass, and v is the velocity of the molecule. Solving for v, we get: \(v = \sqrt{\frac{2 \cdot KE}{m}}\) Now, let's plug in the given values for KE and m: \(v = \sqrt{\frac{2 \cdot (6.2 \cdot 10^{-21} \mathrm{~J})}{(4.7 \cdot 10^{-26} \mathrm{~kg})}}\)
02

Calculate the initial velocity

By plugging in the values, we can now solve for v: \(v \approx \sqrt{\frac{2 \cdot (6.2 \cdot 10^{-21})}{(4.7 \cdot 10^{-26})}} \approx 391.62 \mathrm{~m}/\mathrm{s}\) The initial velocity of the nitrogen molecule is approximately 391.62 m/s.
03

Conservation of energy

We will use the conservation of energy principle, which states that the total mechanical energy (the sum of kinetic and potential energy) remains constant in the absence of any external forces. Let's call the maximum height h: At the starting point (sea level): KE = \(6.2 \cdot 10^{-21}\) J, PE = 0 At maximum height: KE = 0, PE = mgh Since the total energy remains constant, we can equate the sum of the kinetic and potential energy at the beginning to the sum at the maximum height of the molecule: \(KE_{initial} + PE_{initial} = KE_{final} + PE_{final}\)
04

Calculate the height (h)

By plugging in the known values and solving for h, we get: \((6.2 \cdot 10^{-21} \mathrm{~J}) = 0 + (4.7 \cdot 10^{-26} \mathrm{~kg}) \cdot (9.81 \mathrm{~m}/\mathrm{s^2}) \cdot h\) Solve for h: \(h \approx \frac{(6.2 \cdot 10^{-21})}{(4.7 \cdot 10^{-26}) \cdot 9.81} \approx 1.346 \cdot 10^{5} \mathrm{~m}\) The nitrogen molecule would rise up to approximately 1.346 x 10^5 m high.
05

Calculate the percentage of the Earth's radius

The Earth's radius is approximately 6,371,000 meters. To find the percentage of the Earth's radius of this height, we can calculate the ratio between the maximum height and the Earth's radius: \(percentage \approx \frac{1.346 \cdot 10^{5} \mathrm{~m}}{6,371,000 \mathrm{~m}} \cdot 100\) Now let's calculate the percentage: \(percentage \approx \frac{1.346 \cdot 10^{5}}{6,371,000} \cdot 100 \approx 2.114 \cdot 10^{-2}\%\) This height represents approximately 0.02114% of the Earth's radius. Summary: - The height the nitrogen molecule would rise is approximately 1.346 x 10^5 m. - This height represents about 0.02114% of the Earth's radius. - The initial speed of the molecule is approximately 391.62 m/s.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is an essential concept in physics. It states that the total energy in a closed system remains constant if no external forces act upon it.
This means that energy can neither be created nor destroyed; it can only be transformed from one form to another.
In the context of our exercise, we're dealing with mechanical energy, which consists of kinetic energy (KE) and potential energy (PE).

Initially, a nitrogen molecule at sea level has a certain amount of kinetic energy and no potential energy, as it is at the reference height of zero.
As the molecule rises, its height increases, converting kinetic energy into potential energy.
At the maximum height, all the kinetic energy is converted into potential energy.
  • Initial point: KE = 6.2 x 10^{-21} J, PE = 0 J
  • Maximum height: KE = 0 J, PE = mgh
By conserving the total mechanical energy, we equate the initial kinetic energy with the potential energy at the highest point. This enables us to calculate how high the molecule would rise before its energy runs out and it starts to fall back down.
Potential Energy
Potential energy (PE) is the energy stored by an object due to its position relative to a reference point, often in a gravitational field.
It is given by the formula: \( PE = mgh \), where:
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity (9.81 m/s² at Earth's surface),
  • \(h\) is the height of the object above the reference point.
This form of energy plays a crucial role in processes involving vertical motion.
As an object gains height, its potential energy increases while kinetic energy decreases.
In our example, a nitrogen molecule reaches a point where all its initial kinetic energy is converted to potential energy. By applying the formula for potential energy, we solve for \(h\), demonstrating how high the molecule can rise given its initial energy at sea level.
Earth's Radius
Understanding Earth's radius helps contextualize the scale of distances and heights in scientific calculations.
The average radius of Earth is approximately 6,371 kilometers or 6,371,000 meters.
Knowing the Earth's radius allows us to compare vertical distances to more familiar measurements of planetary scale.

In the exercise, calculating the height to which the nitrogen molecule would rise provides insight into how small this height is relative to Earth's size.
We determined that the height is about 134,600 meters, or approximately 0.02114% of Earth's radius.
This helps us appreciate the sheer size of Earth compared to the relatively short distances molecules travel in such scenarios.

Understanding these measurements can broaden our perspective on the vastness of the planet, enhancing not only scientific comprehension but also enriching our appreciation of space and scale.

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Most popular questions from this chapter

An advertisement claims that a certain \(1200-\mathrm{kg}\) car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in \(8.0 \mathrm{~s}\). What average power must the motor supply in order to cause this acceleration? Ignore losses due to friction.

5.60 A man throws a rock of mass \(m=0.325 \mathrm{~kg}\) straight up into the air. In this process, his arm does a total amount of work \(W_{\text {net }}=115 \mathrm{~J}\) on the rock. Calculate the maximum distance, \(h\), above the man's throwing hand that the rock will travel.

A skydiver is subject to two forces: gravity and air resistance. Falling vertically, she reaches a constant terminal speed at some time after jumping from a plane. Since she is moving at a constant velocity from that time until her chute opens, we conclude from the work-kinetic energy theorem that, over that time interval, a) the work done by gravity is zero. b) the work done by air resistance is zero. c) the work done by gravity equals the negative of the work done by air resistance. d) the work done by gravity equals the work done by air resistance. e) her kinetic energy increases.

A car of mass \(m\) accelerates from rest along a level straight track, not at constant acceleration but with constant engine power, \(P\). Assume that air resistance is negligible. a) Find the car's velocity as a function of time. b) A second car starts from rest alongside the first car on the same track, but maintains a constant acceleration. Which car takes the initial lead? Does the other car overtake it? If yes, write a formula for the distance from the starting point at which this happens. c) You are in a drag race, on a straight level track, with an opponent whose car maintains a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2} .\) Both cars have identical masses of \(1000 . \mathrm{kg} .\) The cars start together from rest. Air resistance is assumed to be negligible. Calculate the minimum power your engine needs for you to win the race, assuming the power output is constant and the distance to the finish line is \(0.250 \mathrm{mi}\)

Think about the answers to these questions next time you are driving a car: a) What is the kinetic energy of a \(1500 .-\mathrm{kg}\) car moving at \(15.0 \mathrm{~m} / \mathrm{s} ?\) b) If the car changed its speed to \(30.0 \mathrm{~m} / \mathrm{s}\), how would the value of its kinetic energy change?

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