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Chapter 1: Problem 49

A traditional unit of length in Japan is the ken \((1 \mathrm{ken}=1.97 \mathrm{~m})\). What are the ratios of (a) square kens to square meters and (b) cubic kens to cubic meters? What is the volume of a cylindrical water tank of height 5.50 kens and radius 3.00 kens in (c) cubic kens and (d) cubic meters?

Short Answer

Expert verified
(a) 1 ken虏 = 3.8809 m虏; (b) 1 ken鲁 = 7.66546 m鲁; (c) 155.42 cubic kens; (d) 1188.02 cubic meters.

Step by step solution

01

Convert Square Kens to Square Meters

To find the ratio of square kens to square meters, we first square the conversion factor of kens to meters. Since 1 ken = 1.97 meters, we have:\[1 \text{ ken}^2 = (1.97 \text{ m})^2 = 1.97 \times 1.97 \text{ m}^2 = 3.8809 \text{ m}^2\]Thus, 1 square ken equals 3.8809 square meters.
02

Convert Cubic Kens to Cubic Meters

To find the ratio of cubic kens to cubic meters, we cube the conversion factor. Therefore, from 1 ken = 1.97 meters, we calculate:\[1 \text{ ken}^3 = (1.97 \text{ m})^3 = 1.97 \times 1.97 \times 1.97 \text{ m}^3 = 7.66546 \text{ m}^3\]So, 1 cubic ken equals 7.66546 cubic meters.
03

Calculate Volume in Cubic Kens

The volume of a cylindrical tank is given by the formula:\[V = \pi r^2 h\]where \(r\) is the radius and \(h\) is the height. Using kens for both radius and height (3.00 kens and 5.50 kens respectively), the volume is:\[V_{kens} = \pi (3.00)^2 (5.50) = \pi (9.00) (5.50) = 49.5\pi \approx 155.42 \text{ cubic kens}\]
04

Calculate Volume in Cubic Meters

Substitute the dimensions of the tank from kens to meters using the conversion \(1 \text{ ken} = 1.97 \text{ m}\). Thus:- Radius in meters: \(3.00 \text{ kens} \times 1.97 \text{ m/ken} = 5.91 \text{ meters}\)- Height in meters: \(5.50 \text{ kens} \times 1.97 \text{ m/ken} = 10.835 \text{ meters}\)The volume is:\[V_{meters} = \pi (5.91)^2 (10.835) \approx \pi \times 34.88 \times 10.835 \approx 1188.02 \text{ cubic meters}\]

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Key Concepts

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

Japanese Units of Measurement
Japan uses a traditional system of measurement that includes the 'ken' for length. This system is old and was used long before the adoption of the metric system.
In the context of our exercise, 1 ken is equivalent to approximately 1.97 meters.
This knowledge is crucial for conversion during unit conversion exercises.
When working with these units, it's key to understand that:
  • 1 ken is 1.97 meters of length.
  • 1 square ken is the area represented by a square with sides of 1 ken each.
  • 1 cubic ken is the volume of a cube with sides of 1 ken each.
    These conversions are vital when calculating or comparing measurements in these units to metric units for practical problems.
Volume Calculation
Understanding how volume works is essential for various disciplines, from physics to engineering. Volume is a measure of how much space an object occupies.
To calculate the volume of a regular shape, such as a cylinder, specific formulas are needed:
  • The volume formula for a cylinder is given by: \(V = \pi r^2 h\)
  • Here, \(r\) is the radius, and \(h\) is the height of the cylinder.
  • The radius is the distance from the center of the base to its edge, and the height is the perpendicular distance between the base and the top.
    Plug the values into the equation to find the volume in respective units.
    Remember, using appropriate unit conversions here is essential, such as converting kens to meters if needed for further calculations or interpretations.
Cylindrical Volume
A cylinder is a 3D shape with two identical flat ends, like a can. Calculating its volume helps understand how much liquid it can hold or the amount of material it can contain.
To find the volume of a cylinder using kens as a unit:
  • Use the formula \(V = \pi r^2 h\)
  • Substitute the given radius and height in kens.
  • Calculate \(V\) in cubic kens, using \(\pi\) which is approximately 3.14159.
    To convert this into cubic meters, you must convert the dimensions into meters and recalculate using the same formula. This will give the measurement in cubic meters.
Understanding these units for cylindrical volume calculations helps in practical scenarios like tank and reservoir design.

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Most popular questions from this chapter

A tourist purchases a car in England and ships it home to the United States. The car sticker advertised that the car's fuel consumption was at the rate of 40 miles per gallon on the open road. The tourist does not realize that the U.K. gallon differs from the U.S. gallon: $$ \begin{aligned} 1 \mathrm{U} . \mathrm{K} . \text { gallon } &=4.5460900 \text { liters } \\ 1 \mathrm{U.S} . \text { gallon } &=3.7854118 \text { liters. } \end{aligned} $$ For a trip of 750 miles (in the United States), how many gallons of fuel does (a) the mistaken tourist believe she needs and (b) the car actually require?

(a) Assuming that water has a density of exactly \(1 \mathrm{~g} / \mathrm{cm}^{3}\), find the mass of one cubic meter of water in kilograms. (b) Suppose that it takes \(10.0 \mathrm{~h}\) to drain a container of \(5700 \mathrm{~m}^{3}\) of water. What is the "mass flow rate," in kilograms per second, of water from the container?

Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 min on a town of area \(26 \mathrm{~km}^{2}\). What volume of water, in acre- feet, fell on the town?

A vertical container with base area measuring \(14.0 \mathrm{~cm}\) by \(17.0 \mathrm{~cm}\) is being filled with identical pieces of candy, each with a volume of \(50.0 \mathrm{~mm}^{3}\) and a mass of \(0.0200 \mathrm{~g}\). Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of \(0.250 \mathrm{~cm} / \mathrm{s}\) at what rate (kilograms per minute) does the mass of the candies in the container increase?

A person on a diet might lose \(2.3 \mathrm{~kg}\) per week. Express the mass loss rate in milligrams per second, as if the dieter could sense the second- by-second loss.
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