{{tag>conversions speed energy power chapter1_1}}
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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Conversions: Speed, Energy, and Power                   #@TaskText_HTML@#   

Convert the following values step by step: 

1.  A vehicle speed of $80.00~{\rm {km}\over{h}}$ in ${\rm {m}\over{s}}$

#@HiddenBegin_HTML~f1, Fast Solution~@#
\begin{align*}
v = 80.00~{\rm {km}\over{h}} \qquad \rightarrow \qquad {{\rm {numerical~value}}\over{3.6}} \qquad \rightarrow \qquad  
v = 22.22~{\rm { m}\over{s}}
\end{align*}
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#@PathBegin_HTML~1~@#
\begin{align*}
v &= 80.00~{\rm {km}\over{h}} = 80~{\rm {1'000~m}\over{3'600~s}} = 80~{\rm {m}\over{3.6~s}} \\
  &= 22.22~{\rm {m}\over{s}}
\end{align*}
#@PathEnd_HTML@#

#@ResultBegin_HTML~1~@#
\begin{align*}
v = 22.22~{\rm {m}\over{s}}
\end{align*}
#@ResultEnd_HTML@#

2. An energy of $60.0~{\rm J}$ in ${\rm kWh}$ ($1~{\rm J} = 1~{\rm Joule} = 1~{\rm Watt}\cdot {\rm second}$).

#@PathBegin_HTML~2~@#
\begin{align*}
P = 60.0~{\rm J} = 60.0~{\rm Ws} = 0.06~{\rm kWs} = 0.0000167~{\rm kWh}= 1.67 \cdot 10^{-5}~{\rm kWh}
\end{align*}
#@PathEnd_HTML@#

#@ResultBegin_HTML~2~@#
\begin{align*}
P = 1.67 \cdot 10^{-5}~{\rm kWh}
\end{align*}
#@ResultEnd_HTML@#

3. The number of electrolytically deposited single positively charged copper ions of $1.2 ~\rm Coulombs$ (a copper ion has the charge of about $1.6 \cdot 10^{-19}{\rm C}$)

#@PathBegin_HTML~3~@#
\begin{align*}
n_{\rm ions}= {{{\rm charge}}\over{{\rm charge\,per\,ion}}}={{1.2~{\rm C}}\over{1.6\cdot10^{-19}~{\rm C/ion}}}=7.5 \cdot 10^{18}~{\rm ions}
\end{align*}
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#@ResultBegin_HTML~3~@#
\begin{align*}
n_{\rm ions}= 7.5 \cdot 10^{18}~{\rm ions}
\end{align*}
#@ResultEnd_HTML@#

4. Absorbed energy of a small IoT consumer, which consumes $1~{\rm µW}$ uniformly in $10 ~\rm days$ 

#@PathBegin_HTML~4~@#
\begin{align*}
10 ~{\rm days} &= 10 \cdot {{24 ~\rm h}\over{1 ~\rm day}} \cdot {{60 ~\rm min}\over{1 ~\rm h}} \cdot {{60 ~\rm s}\over{1 ~\rm min}} \\
               &= 10 \cdot   24                           \cdot   60                           \cdot   60~{\rm s} \\
               &= 864'000 ~{\rm s} \\ \\
W_{\rm absorbed} &= 1~{\rm µW} \cdot 864'000~{\rm s} \\
                &= 864'000~{\rm µWs} \\
                &= 864~{\rm mWs}     \\
                &= 0.864~{\rm Ws}    \\
                &= 0.864~{\rm J}     
\end{align*}
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#@ResultBegin_HTML~4~@#
\begin{align*}
W_{\rm absorbed} = 0.864~{\rm Ws} = 0.864~{\rm J}
\end{align*}
#@ResultEnd_HTML@#

#@TaskEnd_HTML@#