Q45P - Fundamentals Of Physics

Question

A unit of time sometimes used in microscopic physics is the shake. One shake equals ${}^{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}$ s. Are there more shakes in a second than there are seconds in a year? (b) Humans have existed for about ${}^{{\mathbf{10}}^{\mathbf{6}}}$ years, whereas the universe is about ${}^{{\mathbf{10}}^{\mathbf{10}}}$ years old. If the age of the universe is defined as${}^{\mathbf{1}}$ “universe day,” where a universe day consists of “universe seconds” as a normal day consists of normal seconds, how many universe seconds have humans existed?

Step-by-Step Solution

Answer

1. Yes, there are more shakes per second than there are seconds per year.

2. Humans existed for ${}^{\mathbf{8}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{universe}\mathbf{‐}\mathbf{sec}}$.

Step 1: Given data

One shake equal ${}^{{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}$.

Humans have existed for about ${}^{{\mathbf{10}}^{\mathbf{6}\mathbf{ }}\mathbf{years}}$.

The universe is about ${}^{{\mathbf{10}}^{\mathbf{10}}\mathbf{ }\mathbf{years}}$ old

Step 2: Understanding the unit conversion

The density of water is${}^{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}\mathbf{ }}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}}$.

Step 3: Understanding the unit conversion

Unit conversion is the conversion between different units of measure for the same quantity. Use the conversion factors to convert the number of shakes in the year and compare it with the number of seconds in the year.

Step 4: (a) Finding the number of seconds in one year

The number of shakes in a second is${}^{{\mathbf{10}}^{\mathbf{8}}}$ .

The number of seconds in years is,

$\mathbf{365}\mathbf{.}\mathbf{25}\mathbf{ }\frac{\mathbf{day}}{\mathbf{year}}\mathbf{\times }\mathbf{24}\frac{\mathbf{hrs}}{\mathbf{day}}\mathbf{\times }\mathbf{60}\frac{\mathbf{min}}{\mathbf{hrs}}\mathbf{\times }\mathbf{60}\frac{\mathbf{sec}}{\mathbf{mins}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{15}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{ }\mathbf{sec}$

Thus, there are more shakes per second than seconds per year.

Step 5: (b) Finding the number of universe seconds of human existence

The age of the universe is,

$\mathbf{1}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{day}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{86400}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{sec}$

The time duration ${}^{\mathbf{\Delta }\mathbf{t}}$  for which humans have existed is calculated as: 

  $$\begin{gathered} \mathbf{\Delta t}\mathbf{=}\frac{{\mathbf{10}}^{\mathbf{6}}}{{\mathbf{10}}^{\mathbf{10}}}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{day} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{day} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{day}\mathbf{\times }\frac{\mathbf{86400}\mathbf{ }\mathbf{u}\mathbf{-}\mathbf{sec}}{\mathbf{1}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{day}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{sec} \end{gathered}$$

Thus, the time duration for which humans have existed is ${}^{\mathbf{8}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{u}\mathbf{‐}\mathbf{sec}}$.