/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 (I) At room temperature, an oxyg... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(I) At room temperature, an oxygen molecule, with mass of 5.31 \(\times\)10\(^{-26}\) kg, typically has a kinetic energy of about 6.21 \(\times\) 10\(^{-21}\) J. How fast is it moving?

Short Answer

Expert verified
The oxygen molecule is moving at approximately 484.7 m/s.

Step by step solution

01

Understanding the Problem

We need to find the velocity of the oxygen molecule given its mass and kinetic energy. The relationship for kinetic energy is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( KE \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity.
02

Rearranging the Formula

To solve for velocity \( v \), we need to rearrange the kinetic energy formula: \( v = \sqrt{\frac{2 \times KE}{m}} \).
03

Substituting the Values

Substitute the given values into the rearranged formula: \( v = \sqrt{\frac{2 \times 6.21 \times 10^{-21} \text{ J}}{5.31 \times 10^{-26} \text{ kg}}} \).
04

Calculating the Velocity

Perform the calculation inside the square root first, and then take the square root to find \( v \). This gives \( v = \sqrt{\frac{1.242 \times 10^{-20}}{5.31 \times 10^{-26}}} = \sqrt{2.34 \times 10^5} \approx 484.7 \text{ m/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Velocity Calculation
Velocity is a measure of how fast something is moving in a certain direction. In physics, when we talk about velocity, we are often dealing with objects influenced by forces. To calculate the velocity of an object like an oxygen molecule, we can use the concept of kinetic energy.

Let's begin with the kinetic energy formula:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \( KE \) is the kinetic energy, \( m \) is the mass, and \( v \) is velocity. Our goal is to solve for \( v \).

To do this, rearrange the formula to isolate \( v \) on one side. This gives us:
  • \( v = \sqrt{\frac{2 \times KE}{m}} \)
Once you have this formula, plugging in the values for mass and kinetic energy allows you to find the velocity with straightforward calculations.
Molecular Motion
Molecules are constantly in motion. This movement is the essence of what we call molecular motion. Understanding this motion is important because it’s one of the key reasons why materials have different states, like solids, liquids, and gases. In our exercise, the oxygen molecule in question is a great example of molecular motion at work.

The velocity of a molecule can tell us a lot about its state and energy. At room temperature, molecules like oxygen are moving around at significant speeds. It's the kinetic energy that results from this molecular motion that we're calculating. This energy is influenced by both the mass of the molecule and how fast it’s moving, which is why molecules of different sizes move at different speeds under the same conditions.

Moreover, this motion is directly related to temperature. An increase in temperature generally means an increase in the kinetic energy and velocity of the molecules. Therefore, molecular motion gives us insights into physical properties and behaviors of substances.
Kinetic Energy Formula
The kinetic energy formula is a fundamental equation in physics, especially when dealing with moving objects. Kinetic energy (
  • \( KE \)
) quantifies the energy of motion. It is given by the equation:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \( m \) is the mass, and \( v \) is the velocity of the moving object.

The formula indicates that kinetic energy is directly proportional to both the mass of an object and the square of its velocity. This means that even slight changes in velocity result in significant changes in kinetic energy. This quadratic relationship emphasizes the importance of speed in determining an object's kinetic energy.

In practical scenarios like the one in the exercise, the kinetic energy formula helps us calculate how fast an object is moving by having either the mass or kinetic energy given and determining the missing value. This ties back into how we understand movement and energy interactions at the molecular level and in larger bodies.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Many cars have "5 mi/h (8 km/h) bumpers" that are designed to compress and rebound elastically without any physical damage at speeds below 8 km/h. If the material of the bumpers permanently deforms after a compression of 1.5 cm, but remains like an elastic spring up to that point, what must be the effective spring constant of the bumper material, assuming the car has a mass of 1050 kg and is tested by ramming into a solid wall?

(I) How much work must be done to stop a 925-kg car traveling at 95 km/h?

A cyclist starts from rest and coasts down a 4.0\(^\circ\) hill. The mass of the cyclist plus bicycle is 85 kg. After the cyclist has traveled 180 m, (\(a\)) what was the net work done by gravity on the cyclist? (\(b\)) How fast is the cyclist going? Ignore air resistance and friction.

(II) A 380-kg piano slides 2.9 m down a 25\(^\circ\) incline and is kept from accelerating by a man who is pushing back on it \(parallel\) \(to\) \(the\) \(incline\) (Fig. 6-36). Determine: (\(a\)) the force exerted by the man, (\(b\)) the work done on the piano by the man, (\(c\)) the work done on the piano by the force of gravity, and (\(d\)) the net work done on the piano. Ignore friction.

(III) A cyclist intends to cycle up a 7.50\(^\circ\) hill whose vertical height is 125 m. The pedals turn in a circle of diameter 36.0 cm. Assuming the mass of bicycle plus person is 75.0 kg, (\(a\)) calculate how much work must be done against gravity. (\(b\)) If each complete revolution of the pedals moves the bike 5.10 m along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.