/*! 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 93 A physics student spends part of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 \(\mathrm{W}\) . The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 \(\mathrm{W}\) . If she expends a total of \(1.1 \times 10^{7} \mathrm{J}\) of energy in a 24 -hour day, how much of the day did she spend walking?

Short Answer

Expert verified
She spent approximately 3.64 hours walking.

Step by step solution

01

Define Variables

Let \( t_w \) be the number of hours spent walking. For the rest of the day, she spends \( 24 - t_w \) hours resting.
02

Set Up the Equation

The total energy expenditure can be expressed as:\[280t_w + 100(24 - t_w) = 1.1 \times 10^7\]where both terms on the left represent the energy spent walking and resting, respectively, in joules.
03

Convert Hours to Seconds

Since the rates are in watts (Joules per second), we need to convert hours to seconds by multiplying by 3600 (the number of seconds in an hour). Thus, the equation becomes:\[280 \times t_w \times 3600 + 100 \times (24-t_w) \times 3600 = 1.1 \times 10^7\]
04

Simplify the Equation

Simplify the equation:\[280 \times 3600 t_w + 100 \times 3600 \times 24 - 100 \times 3600 t_w = 1.1 \times 10^7\]This simplifies to:\[1008000t_w + 8640000 - 360000t_w = 1.1 \times 10^7\]
05

Solve for \( t_w \)

Combine like terms:\[648000t_w + 8640000 = 1.1 \times 10^7\]Subtract 8640000 from both sides:\[648000t_w = 1.1 \times 10^7 - 8640000\]\[648000t_w = 2360000\]Divide both sides by 648000 to find \( t_w \):\[t_w = \frac{2360000}{648000}\]\[t_w \approx 3.64\]
06

Convert the Answer to Hours

Since the calculation is based on hours, and we solved for \( t_w \), the time she spent walking is \( 3.64 \) hours.

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.

Physics Problem Solving
Physics problems can sometimes seem tricky, but breaking them down into manageable steps can help demystify the process. The key to solving any physics problem is understanding what you're being asked. Carefully read the question to identify what data is provided and what you need to calculate.
In our exercise, we started by defining the variables. Here, the total time is 24 hours (a full day), and we need to find out how many of those hours were spent walking. Establishing clear variables such as the time spent walking and resting allows us to set up equations that represent the real-world scenario.
Building equations from word problems requires identifying relationships between variables, often done through the lens of physics principles like conservation laws or known formulas. Once the equation is set up, the next step involves manipulating algebraic equations—combining like terms, isolating the unknown variable, and using arithmetic operations to simplify and solve the equations. This process is a central part of problem solving in physics.
Unit Conversion in Physics
Unit conversion is often necessary in physics to ensure consistency across the problem. Units need to match across all terms in an equation to perform meaningful calculations. In many physics problems, energy is given in joules, while rates of energy expenditure are provided in watts (Joules per second). This poses the challenge of converting units to ensure calculations are coherent.
Our original exercise solution involved converting hours into seconds because the power unit, watts, translates to Joules per second. Since there are 3600 seconds in an hour, multiplying the number of hours by 3600 provides the equivalent time in seconds. This conversion is crucial because it aligns with the rate of energy expenditure given in watts, allowing for accurate calculation of total energy in joules.
Always remember, consistency is key in physics. When encountering different units, identifying a common base unit—be it time, length, mass, or others—facilitates the solving process, ensuring that our results are accurate and meaningful. Unit alignment prevents errors and miscalculations that could arise from incompatible unit operations.
Work and Energy
The concepts of work and energy are central to physics and encompass a variety of everyday phenomena. Work is done when a force causes an object to move, and energy is often referred to as the capacity to do work. In physics, these concepts are often interconnected and quantified in terms of power, a measure of how quickly work is done or energy is transferred.
In our exercise, the energy expenditure rates provided in watts signify how much energy is being used per second. The problem combines these rates to determine the total work—energy expended—over the course of a day. This daily energy expenditure is calculated using the formula for work and energy, where power (in watts) is multiplied by time (in seconds) to compute energy in joules.
Understanding work and energy within the context of their mathematical relationships is vital. The rearrangement and application of the formula \[ P = \frac{W}{t} \] where \( P \) is power, \( W \) is work or energy, and \( t \) is time, illustrate these concepts in practical scenarios, showing not just theoretical understanding, but real-life applications of physics principles. This foundational knowledge underpins many branches of physics, making it essential for students to grasp thoroughly.

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

A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{g},\) and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{\imath}-(40 \mathrm{N}) \hat{\mathrm{J}}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\boldsymbol{\imath}}-(3.0 \mathrm{m}) \hat{\boldsymbol{J}}\) . How much work does the force you apply do on the grocery cart?

CALC A force in the \(+x\) -direction with magnitude \(F(x)=18.0 \mathrm{N}-(0.530 \mathrm{N} / \mathrm{m}) x\) is applied to a 6.00 -kg box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x=0,\) what is its speed after it has traveled 14.0 \(\mathrm{m}\) ?

BIO All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 \(\mathrm{g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 70 -kg athlete can maintain a power output of 1.4 \(\mathrm{kW}\) for no more than a few seconds; the steady power output of a typical athlete is only 500 \(\mathrm{W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

You and your bicycle have combined mass 80.0 \(\mathrm{kg} .\) When you reach the bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 6.78)\) . At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s} .\) You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

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.