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Mount Everest, at 29,028 ft, is the tallest mountain on Earth. What is its height in kilometers? (Assume that \(1 \mathrm{m}=\) 3.281 ft.)

Short Answer

Expert verified
The height of Mount Everest is approximately 8.848 km.

Step by step solution

01

Convert Feet to Meters

The exercise mentions that 1 meter equals 3.281 feet. Therefore, to convert the height of Mount Everest to meters, divide its height in feet, which is 29,028 ft, by 3.281 ft/m. This is represented as follows: \(29,028 ft \div 3.281 ft/m\).
02

Convert Meters to Kilometers

Next, convert the result from the previous step from meters to kilometers. Remember that 1 kilometer equals 1000 meters, so to convert meters to kilometers, divide the number of meters by 1000 m/km. This is represented as follows: \((29,028 ft \div 3.281 ft/m) \div 1000 m/km\).
03

Calculation

Now, perform the division operations in the indicated order to find the height of Mount Everest in kilometers.

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Key Concepts

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

Convert Feet to Meters
When tackling problems in physical sciences, one frequently encounters the need to convert measurements from one unit to another. For instance, converting feet to meters is a common task in physics and international contexts where the metric system prevails. So, how do we go about this?

Feet and meters are units used to measure length. One foot is defined as 0.3048 meters. Therefore, to convert feet to meters, one multiplies the number of feet by 0.3048. The formula used is: \

\[\text{meters} = \text{feet} \times 0.3048\]

In the context of our Mount Everest exercise, to convert the mountain's height from feet to meters, we take the height in feet (\(29,028\) ft) and multiply it by 0.3048. This straightforward calculation will give us the height in meters, which is the first step to obtaining the measurement in the metric system's unit for longer distances: kilometers.
Metric System
The metric system is an international system of measurement that is based on powers of ten. It was designed to provide a standard for weights and measures around the world. In the metric system, length is measured in meters, weight in kilograms, and volume in liters. This system simplifies mathematical and scientific calculations by using a base unit and multiplying it by powers of ten to signify larger or smaller quantities.

For length, the fundamental unit is the meter. Larger and smaller lengths are then expressed in terms of meters using prefixes like kilo (thousand), centi (hundredth), and milli (thousandth). For example, a kilometer is 1000 meters, whereas a centimeter is one-hundredth of a meter, and a millimeter is one-thousandth of a meter. When working in the metric system, converting between units becomes a straightforward process of moving the decimal point.
Dimensional Analysis
Dimensional analysis is a powerful tool used in physics and engineering to convert one set of measurements to another. It involves the use of conversion factors and the principle that units of measurement can be treated as mathematical quantities that can cancel each other out.

This technique helps to ensure that equations make sense dimensionally. If one side of an equation deals with distance, the other side must also relate to distance. For conversions, dimensional analysis uses the fact that one measurement unit divided by itself is equal to one, and therefore, a quantity remains unchanged when it is multiplied by one.

A conversion factor is essentially a fraction equivalent to one. For example, given that 3.281 feet equals one meter, the conversion factors would be \[\frac{1 \, \text{m}}{3.281 \, \text{ft}}\] or its reciprocal \[\frac{3.281 \, \text{ft}}{1 \, \text{m}}\]. When you multiply a measurement by a conversion factor, you ensure that you only change the units, not the physical quantity being measured. In the Mount Everest example, the use of dimensional analysis allows a seamless transition from feet to meters and ultimately to kilometers, ensuring the final answer is both numerically correct and dimensionally consistent.

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Most popular questions from this chapter

The first atomic bomb was detonated on July 16,1945 at the Trinity test site about \(200 \mathrm{mi}\) south of Los Alamos. In 1947 , the U.S. government declassified a film reel of the explosion. From this film reel, British physicist G. I. Taylor was able to determine the rate at which the radius of the fireball from the blast grew. Using dimensional analysis, he was then able to deduce the amount of energy released in the explosion, which was a closely guarded secret at the time. Because of this, Taylor did not publish his results until 1950. This problem challenges you to recreate this famous calculation. (a) Using keen physical insight developed from years of experience, Taylor decided the radius \(r\) of the fireball should depend only on time since the explosion, \(t,\) the density of the air, \(\rho,\) and the energy of the initial explosion, \(E .\) Thus, he made the educated guess that \(r=k E^{a} \rho^{b} t^{c}\) for some dimensionless constant \(k\) and some unknown exponents \(a, b,\) and \(c .\) Given that \([\mathrm{E}]=\mathrm{ML}^{2} \mathrm{T}^{-2}\) determine the values of the exponents necessary to make this equation dimensionally consistent. (Hint: Notice the equation implies that \(k=r E^{-a} \rho^{-b} t^{-c}\) and that \([k]=1\) ) (b) By analyzing data from high-energy conventional explosives, Taylor found the formula he derived seemed to be valid as long as the constant \(k\) had the value \(1.03 .\) From the film reel, he was able to determine many values of \(r\) and the corresponding values of \(t .\) For example, he found that after \(25.0 \mathrm{ms}\), the fireball had a radius of \(130.0 \mathrm{m}\). Use these values, along with an average air density of 1.25 \(\mathrm{kg} / \mathrm{m}^{3},\) to calculate the initial energy release of the Trinity detonation in joules (J). (Hint: To get energy in joules, you need to make sure all the numbers you substitute in are expressed in terms of SI base units.) (c) The energy released in large explosions is often cited in units of "tons of TNT" (abbreviated "t TNT"), where 1 t TNT is about 4.2 GJ. Convert your answer to (b) into kilotons of TNT (that is, kt TNT). Compare your answer with the quickand-dirty estimate of \(10 \mathrm{kt}\) TNT made by physicist Enrico Fermi shortly after witnessing the explosion from what was thought to be a safe distance. (Reportedly, Fermi made his estimate by dropping some shredded bits of paper right before the remnants of the shock wave hit him and looked to see how far they were carried by it.)

Can the validity of a model be limited or must it be universally valid? How does this compare with the required validity of a theory or a law?

American football is played on a 100 -yd-long field, excluding the end zones. How long is the field in meters? (Assume that \(1 \mathrm{m}=3.281 \mathrm{ft}\) )

The arc length formula says the length \(s\) of arc subtended by angle \(\sim\) in a circle of radius \(r\) is given by the equation \(s=r \Theta .\) What are the dimensions of (a) \(s\) (b) \(r,\) and \((\mathrm{c}) \Theta ?\)

Soccer fields vary in size. A large soccer field is 115 \(\mathrm{m}\) long and \(85.0 \mathrm{m}\) wide. What is its area in square feet? (Assume that \(1 \mathrm{m}=3.281 \mathrm{ft}\) )

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