Verified Commit 9053814a authored by Soren's avatar Soren

Make some fixes to Physics

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#+TITLE: Review
#+DATE: <2018-08-29 Wed>
#+AUTHOR: Soren
#+EMAIL: soren@gallium
#+EMAIL: soren@disroot.org
#+LANGUAGE: en
#+SELECT_TAGS: export
#+EXCLUDE_TAGS: noexport
#+CREATOR: Emacs 25.2.2 (Org mode 9.1.14)
#+LaTeX_HEADER: \usepackage{siunitx}
* Section Review [1.2]
......@@ -18,7 +19,7 @@ The ruler can be placed or supported more effectively.
** Tools [19]
You find a micrometer (a tool used to measure objects to the
nearest 0.01 mm) that has been badly bent. How would it compare
nearest $0.01 \si{mm}$) that has been badly bent. How would it compare
to a new, high quality meterstick in terms of its precision?
Its accuracy?
......@@ -26,12 +27,12 @@ The bent micrometer would be more precise, but less accurate.
** Critical Thinking [23]
Your friend states in a report that the average time to circle a
1.5mi track was 65.414s. This was measured by timing 7 laps using
a clock with a precision of 0.1s. How much confidence do you have
in the results of the report? Explain.
$1.5 \si{mi}$ track was $65.414 \si{s}$. This was measured by timing
7 laps using a clock with a precision of $0.1 \si{s}$. How much
confidence do you have in the results of the report? Explain.
Not much confidence:
1. There is human error in the timing measurements.
2. The data set is too small.
3. Not enough context is given and variables are not explained.
4. 65.414s is too precise for a 0.1s precise timer.
4. $65.414 \si{s}$ is too precise for a $0.1 \si{s}$ precise timer.
#+TITLE: 30-08-2018-Assessment
#+TITLE: Assessment
#+DATE: <2018-08-30 Thu>
#+AUTHOR: Soren
#+EMAIL: soren@gallium
#+EMAIL: soren@disroot.org
#+LANGUAGE: en
#+SELECT_TAGS: export
#+EXCLUDE_TAGS: noexport
#+CREATOR: Emacs 25.2.2 (Org mode 9.1.14)
#+LaTeX_HEADER: \usepackage{siunitx}
* Chapter 1 Assessment
** Mathematics and Physics [67]
Convert each of the following measurements to meters:
- $42.3 \text{cm}$ :: $100 \text{cm} = 1 \text{m}$,
$42.3 \text{cm} = 0.423 \text{m}$.
- $6.2 \text{pm}$ :: $1 \text{pm} = 1000000000000000 \text{m}$,
$6.2 \text{pm} = 6200000000000 \text{m}$.
- $21 \text{km}$ :: $1 \text{km} = 1000 \text{m}$,
$21 \text{km} = 21000 \text{m}$.
- $0.023 \text{mm}$ :: $1000 \text{mm} = 1 \text{m}$,
$0.023 \text{mm} = 0.000023 \text{m}$.
- $214 \text{µm}$ :: $1000000 \text{µm} = 1 \text{m}$,
$214 \text{µm} = 0.000214 \text{m}$.
- $57 \text{nm}$ :: $1000000000 \text{nm} = 1 \text{m}$,
$57 \text{nm} = 0.000000057 \text{m}$.
- $42.3 \si{cm}$ :: $100 \si{cm} = 1 \si{m}$,
$42.3 \si{cm} = \num{4.23d-1} \si{m}$.
- $6.2 \si{pm}$ :: $1 \si{pm} = \num{1d15} \si{m}$,
$6.2 \si{pm} = \num{6.2d15} \si{m}$.
- $21 \si{km}$ :: $1 \si{km} = \num{1d3} \si{m}$,
$21 \si{km} = \num{21d3} \si{m}$.
- $\num{2.3d-2} \si{mm}$ :: $\num{1d3} \si{mm} = 1 \si{m}$,
$\num{2.3d-2} \si{mm} = \num{2.3d-6} \si{m}$.
- $214 \si{µm}$ :: $\num{1d6} \si{µm} = 1 \si{m}$,
$214 \si{µm} = \num{2.14d-5} \si{m}$.
- $57 \si{nm}$ :: $\num{1d9} \si{nm} = 1 \si{m}$,
$57 \si{nm} = \num{5.7d-9} \si{m}$.
** Measurement [75]
A water tank has a mass of 3.64 kg when it is empty and a mass
sf 51.8 kg when it is filled to a certain level. What is the
A water tank has a mass of $3.64 \si{kg}$ when it is empty and a mass
of $51.8 \si{kg}$ when it is filled to a certain level. What is the
mass of the water in the tank?
$51.8 \text{kg} - 3.64 \text{kg} = 48.16 \text{kg}$.
$51.8 \si{kg} - 3.64 \si{kg} = 48.16 \si{kg}$.
** Measurement [79]
Give the measure shown in *Figure 1-24* (Class A scale pointing
......
#+TITLE: Homework
#+DATE: <2018-11-29 Thu>
#+AUTHOR: Soren
#+EMAIL: soren@disroot.org
#+LANGUAGE: en
#+SELECT_TAGS: export
#+EXCLUDE_TAGS: noexport
#+CREATOR: Emacs 26.1 (Org mode 9.1.9)
#+LaTeX_HEADER: \usepackage{siunitx}
* Section Review [6.2]
** [12]
A runner moving at the speed of 8.8 m/s rounds a bend with a radius
of 25 m. What is the centripetal acceleration of the runner, and
what agent exerts force on the runner?
\begin{align}
a &= \frac{V^{2}}{r} \\
a &= \frac{77.44 \si{m^{2}/s^{2}}}{25 \si{m}} \\
a &= 3.1 \si{m/s^{2}}
\end{align}
This acceleration is due to a friction force from the ground.
** [14]
An airplane traveling at $201 \si{m/s}$ makes a turn. What is the
smallest radius of the circular path (in $\si{km}$) that the pilot
can make and keep the centripetal acceleration under
$5.0 \si{m/s^{2}}$?
\begin{align}
r &\geq \frac{V^{2}}{a} \\
r &\geq \frac{40401 \si{m^{2}/s^{2}}}{5 \si{m/s^{2}}} \\
r &\geq 8080.2 \si{m} \\
r &\geq 8.08 \si{km}
\end{align}
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