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Chapter 1: Problem 6

The following conversions occur frequently in physics and are very useful. (a) Use \(1 \mathrm{mi}=5280 \mathrm{ft}\) and \(1 \mathrm{~h}=3600 \mathrm{~s}\) to convert \(60 \mathrm{mph}\) to units of \(\mathrm{ft} / \mathrm{s}\). (b) The acceleration of a freely falling object is \(32 \mathrm{ft} / \mathrm{s}^{2}\). Use \(1 \mathrm{ft}=30.48 \mathrm{~cm}\) to express this acceleration in units of \(\mathrm{m} / \mathrm{s}^{2}\). (c) The density of water is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\). Convert this density to units of \(\mathrm{kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The converted units are: (a) 60 mph = 88 ft/s, (b) 32 ft/s虏 = 9.8 m/s虏, (c) 1.0 g/cm鲁 = 1000 kg/m鲁.

Step by step solution

01

Conversion of Speed

Convert 60 mph into ft/s using given unit conversions: \(1 \mathrm{mi}=5280 \mathrm{ft}\) and \(1 \mathrm{~h}=3600 \mathrm{~s}\). Therefore, \(60 \mathrm{mi/h} = 60 \times \frac{5280\ \mathrm{ft}}{1\ \mathrm{mi}} \times \frac{1\ \mathrm{h}}{3600\ \mathrm{s}} = 88 \mathrm{ft/s}\)
02

Conversion of Acceleration

Convert the acceleration of 32 ft/s虏 into m/s虏 using the given unit conversion: \(1 \mathrm{ft}=30.48 \mathrm{~cm}\). Note: 1 m = 100 cm. Thus, \(32 \frac{\mathrm{ft}}{\mathrm{s}^2} = 32 \times \frac{30.48\ \mathrm{cm}}{1\ \mathrm{ft}} \times \frac{1 \ \mathrm{m}}{100\ \mathrm{cm}} = 9.8 \frac{\mathrm{m}}{\mathrm{s}^2}\).
03

Conversion of Density

Convert the density of water from g/cm鲁 into kg/m鲁. Note that \(1 \mathrm{kg} = 1000 \mathrm{g}\) and \(1 \mathrm{m}=100 \mathrm{cm}\). Therefore, \(1.0 \mathrm{g}/\mathrm{cm}^3 = 1.0 \times \frac{1000 \mathrm{kg}}{1 \mathrm{g}} \times \left(\frac{1 \mathrm{m}}{100 \mathrm{cm}}\right)^3 = 1000 \mathrm{kg}/\mathrm{m}^3\).

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Key Concepts

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

Velocity Conversion
Converting velocity from one set of units to another is a common task in physics. To illustrate, let's look at converting miles per hour (mph) to feet per second (ft/s). Understanding the units and conversion factors is crucial. Here we know that:
  • 1 mile = 5280 feet
  • 1 hour = 3600 seconds
To convert 60 mph to ft/s, multiply 60 by the conversion factors to change miles to feet and hours to seconds: \[60 \text{ mph} = 60 \times \frac{5280\ \text{ft}}{1 \ \text{mi}} \times \frac{1 \ \text{h}}{3600 \ \text{s}} = 88 \text{ ft/s}\] This calculation involves multiplying 60 miles by 5280 to convert miles to feet, and then dividing by 3600 to convert hours to seconds. It's always beneficial to write out the units so they cancel appropriately, making sure the units left over are the ones required.
Acceleration Conversion
Acceleration conversion is key in physics problems, especially when working with gravitational forces. To convert acceleration from feet per second squared (ft/s虏) to meters per second squared (m/s虏), you need to convert the distance units. Remember:
  • 1 foot = 30.48 cm
  • 1 meter = 100 cm
Let's convert 32 ft/s虏 to m/s虏. First, convert feet to centimeters, and then centimeters to meters: \[32 \frac{\text{ft}}{\text{s}^2} = 32 \times \frac{30.48\ \text{cm}}{1\ \text{ft}} \times \frac{1 \ \text{m}}{100\ \text{cm}} = 9.8 \frac{\text{m}}{\text{s}^2}\] The conversion involves multiplying by 30.48 to change ft to cm, and then by \(\frac{1}{100}\) to change cm to m. Writing out the conversion with cancellation helps track units through the calculation.
Density Conversion
Density, often expressed in grams per cubic centimeter (g/cm鲁), can be converted to kilograms per cubic meter (kg/m鲁) using fundamental unit conversion principles. The process involves converting both the mass and volume units:
  • 1 kg = 1000 g
  • 1 m = 100 cm, hence \(\text{1 m}^3 = \text{100}^3 \text{cm}^3\)
Let's convert the density of water, which is 1.0 g/cm鲁, to kg/m鲁: \[1.0 \frac{\text{g}}{\text{cm}^3} = 1.0 \times \frac{1000 \text{ kg}}{1 \text{ g}} \times \left(\frac{1 \text{ m}}{100 \ \text{cm}}\right)^3 = 1000 \frac{\text{kg}}{\text{m}^3}\]The multiplication by 1000 converts grams to kilograms. The cubic conversion from cm鲁 to m鲁 ensures that volume is converted correctly from smaller to larger cubes by applying \(\left(\frac{1}{100}\right)^3\). Visualizing the conversion cancels out the cm鲁 and leaves the required kg/m鲁.

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Most popular questions from this chapter

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about \(500 \mathrm{~cm}^{3}\) of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

A dog in an open field runs \(12.0 \mathrm{~m}\) east and then \(28.0 \mathrm{~m}\) in a direction \(50.0^{\circ}\) west of north. In what direction and how far must the dog then run to end up \(10.0 \mathrm{~m}\) south of her original starting point?

DATA You are a team leader at a pharmaceutical company. Several technicians are preparing samples, and you want to compare the densities of the samples (density = mass/volume) by using the mass and volume values they have reported. Unfortunately, you did not specify what units to use. The technicians used a variety of units in reporting their values, as shown in the following table. $$ \begin{array}{lll} \hline \text { Sample ID } & \text { Mass } & \text { Volume } \\ \hline \text { A } & 8.00 \mathrm{~g} & 1.67 \times 10^{-6} \mathrm{~m}^{3} \\\ \text { B } & 6.00 \mu \mathrm{g} & 9.38 \times 10^{6} \mu \mathrm{m}^{3} \\ \text { C } & 8.00 \mathrm{mg} & 2.50 \times 10^{-3} \mathrm{~cm}^{3} \\ \text { D } & 9.00 \times 10^{-4} \mathrm{~kg} & 2.81 \times 10^{3} \mathrm{~mm}^{3} \\ \text { E } & 9.00 \times 10^{4} \mathrm{ng} & 1.41 \times 10^{-2} \mathrm{~mm}^{3} \\ \text { F } & 6.00 \times 10^{-2} \mathrm{mg} & 1.25 \times 10^{8} \mu \mathrm{m}^{3} \end{array} $$ List the sample IDs in order of increasing density of the sample.

Torque is a vector quantity that specifies the effectiveness of a force in causing the rotation of an object. The torque that a force \(\vec{F}\) exerts on a rigid object depends on the point where the force acts and on the location of the axis of rotation. If \(\vec{r}\) is the length vector from the axis to the point of application of the force, then the torque is \(\vec{r} \times \vec{F}\). If \(\vec{F}\) is \(22.0 \mathrm{~N}\) in the \(-y\) -direction and if \(\vec{r}\) is in the \(x y\) -plane at an angle of \(36^{\circ}\) from the \(+y\) -axis toward the \(-x\) -axis and has magnitude \(4.0 \mathrm{~m},\) what are the magnitude and direction of the torque exerted by \(\overrightarrow{\boldsymbol{F}} ?\)

You are standing on a street corner with your friend. You then travel \(14.0 \mathrm{~m}\) due west across the street and into your apartment building. You travel in the elevator \(22.0 \mathrm{~m}\) upward to your floor, walk \(12.0 \mathrm{~m}\) north to the door of your apartment, and then walk \(6.0 \mathrm{~m}\) due east to your balcony that overlooks the street. Your friend is standing where you left her. Now how far are you from your friend?
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