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Chapter 1: Problem 6

In a new system of units, unit of mass is \(10 \mathrm{~kg}\), unit of length is \(1 \mathrm{~km}\) and unit of time is \(1 \mathrm{~min}\). The value of \(1 \mathrm{~J}\) in this new hypothartical system is (a) \(3.6 \times 10^{-4}\) new units (b) \(6 \times 10^{7}\) new units (c) \(10^{11}\) new units (d) \(1.67 \times 10^{4}\) new units

Short Answer

Expert verified
The value of 1 Joule in the new unit system is \( 3.6 \times 10^{-4} \) new units.

Step by step solution

01

Understand Physical Quantities

The standard units for mass, length, and time in the International System of Units (SI) are kilograms (kg), meters (m), and seconds (s), respectively. In this exercise, we have new units where mass is 10 kg, length is 1 km (1000 m), and time is 1 minute (60 s).
02

Define a Joule in SI Units

In SI units, 1 Joule (J) is defined as 1 kg*m虏/s虏. This is derived from the formula for energy or work: \( E = F \times d = m \times a \times d \), implying \( Joule = \text{mass} \times \text{acceleration} \times \text{distance} \).
03

Convert Mass to New Unit

In the new system, the unit of mass is 10 kg. Therefore, for a mass of 1 kg in this system, the corresponding unit becomes \( \frac{1}{10} \) new mass units.
04

Convert Length to New Unit

The new unit of length is 1 km, which equals 1000 m. Therefore, for a length of 1 m in this system, it becomes \( \frac{1}{1000} \) new length units.
05

Convert Time to New Unit

The new unit of time is 1 minute, equating to 60 seconds. Therefore, for a time of 1 s in this system, the equivalent is \( \frac{1}{60} \) new time units.
06

Convert Joule to New System

Substitute the conversions into the equation for a Joule: \( 1 \text{ J} = 1 \cdot (\text{kg}) \times (\text{m}^2) / (\text{s}^2) = \left( \frac{1}{10} \right) \cdot \left( \frac{1}{1000} \right)^2 / \left( \frac{1}{60} \right)^2 \). Calculate this to find the number of new units for 1 J.
07

Calculation

Compute the expression: \[\left( \frac{1}{10} \right) \cdot \left( \frac{1}{1000} \right)^2 / \left( \frac{1}{60} \right)^2 \rightarrow \frac{1}{10} \times \frac{1}{1000000} \times \frac{3600}{1} = \frac{3600}{10000000} = 3.6 \times 10^{-4} \].

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Key Concepts

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

Hypothetical Unit Systems
In the realm of physics, unit systems play a crucial role for measuring various physical quantities. However, sometimes it becomes necessary to consider hypothetical unit systems. These are systems where the standard base units like mass, length, and time are redefined. Hypothetical unit systems are usually devised to simplify the mathematical treatment of problems or model real-world situations with different known quantities. For instance, if a hypothetical system uses 10 kg for mass, 1 km for length, and 1 minute for time, it alters the way we measure and interact with forces or energy under this new framework. This creative approach can yield convenient units that offer simpler calculations in specific scenarios.
Joule Conversion
Joule is the standard unit for energy in the International System of Units (SI).Its conversion between different unit systems can be complex but extremely insightful when mastered.A joule is defined as the work done by a force of one newton moving an object one meter in the direction of the force.In terms of base SI units, it is expressed as:\[ 1 ext{ J} = 1 ext{ kg} \times 1 ext{ m}^2 / 1 ext{ s}^2 \] Converting a joule requires changing each base unit according to the new system parameters.Like in our hypothetical unit system, for joule conversion, we address the differences in each dimension - mass, length, and time.The process involves substituting new values for these dimensions and solving to express a joule in terms of new units, accurately bridging standard and hypothetical frameworks.
Dimensional Analysis
Dimensional analysis is a powerful and elegant technique in physics.It offers a way to understand and solve equations by considering the base dimensions involved.Dimensions like mass, length, and time help in refining understanding independent of specific units used.When converting units, especially in complex contexts as in hypothetical unit systems, dimensional analysis checks that the units on both sides of an equation are consistent.For example, expressing Joules in a new unit system entails converting each dimension separately:
  • Mass units are analyzed as \( rac{1}{10} \) of the original.
  • Length units as \( rac{1}{1000} \cdot 1000 ext{ m/km} \).
  • Time units as \( rac{1}{60} \cdot 60 ext{ sec/min} \).
Finally, reassemble them into the joule's expression, confirming the dimensional alignment.
Physical Quantities Conversion
The conversion of physical quantities between unit systems is a cornerstone of physics problems-solving. This involves translating numerical values from one set of units to another without affecting the underlying physical phenomenon they represent. To perform a conversion accurately, consider: - The ratio between new and old units for each dimension - The multiplication or division needed to express an entire physical quantity in new units Following this method allows us to compute energy, distance, or force in a manner that aligns with the context-specific unit definitions. In practice, like the exercise above, converting a joule by changing mass, length, and time units into their hypothetical equivalents results in a coherent expression that faithfully translates energy under those new measures.

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Most popular questions from this chapter

The SI unit of length is metre. Suppose we adopt a new unit of length which equal \(x\) metre. The area of \(1 \mathrm{~m}^{2}\) expressed in terms of the new unit has a magnitude (a) \(x\) (b) \(x^{2}\) (c) \(x^{-1}\) (d) \(x^{-2}\)

The work done by a battery is \(W=\varepsilon \Delta q\), where \(\Delta q\) charge transferred by battery, \(\varepsilon=\) emf of the battery. What are dimensions of emf of battery? (a) \(\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\) (b) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\) (c) \(\left[\mathrm{M}^{2} \mathrm{~L}^{0} \mathrm{~T}^{-3} \mathrm{~A}^{0}\right]\) (d) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]\)

The wavelength associated with a moving particle depends upon power \(p\) of its mass \(m, q\) th power of its velocity \(v\) and \(r\) th power of Planck's constant \(h\). Then the correct set of values of \(p, q\) and \(r\) is (a) \(p=1, q=-1, r=1\) (b) \(p=1, q=1, r=1\) (c) \(p=-1, p=-1, r=-1\) (d) \(p=-1, q=-\mathbf{L} r=1\)

Which of the following is the most precise device for measuring length? (a) A vernier callipers with 20 divisions on the sliding scale (b) A screw gauge of pitch \(1 \mathrm{~mm}\) and 100 divisions on the circular scale (c) An optical instrument that can measure length to within a wavelength of light? (d) All are equally precise device for measuring length

An important milestone in the evolution of the universe just after the Big Bang is the Planck time \(t_{p}\), the value of which depends on three fundamental constants speed \(c\) of light in vacuum, gravitational constant \(G\) and Planck's constant \(h\). Then, \(t_{p} \propto\) (a) \(\mathrm{Ghc}^{5}\) (b) \(\frac{c^{5}}{G h}\) (c) \(\frac{G h}{c^{5}}\) (d) \(\left(\frac{G h}{c^{5}}\right)^{1 / 2}\)
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