Updated spacing between @ latex expressions

astrosander правка от 6 окт. 2024 г., 00:48

правка #10276 предыдущая #9269 GitHub c702ebe ← раньше позже →

en/1.3.20.md +3 −3

			<!DOCTYPE html>
			<html lang="en">

			<head>
			<meta charset="utf-8">
			<meta name="viewport" content="width=device-width, initial-scale=1.0">
			<meta http-equiv="content-language" content="en">
			<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
			<meta name="description" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.</title>
			<link rel="stylesheet" href="../../css/css-latex/style.css">
			<link rel="icon" href="../../img/logo.png" type="image/png">
			<script src="../../js/jquery-1.10.1.min.js"></script>
			<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
			<script type="text/x-mathjax-config">
			MathJax.Hub.Config({
			extensions: ['tex2jax.js'],
			jax: ['input/TeX', 'output/HTML-CSS'],
			tex2jax: {
			inlineMath: [['$', '$'], ['$', '$']],
			processEscapes: true,
			processClass: 'tex2jax',
			ignoreClass: 'html'
			},
			showProcessingMessages: false,
			messageStyle: 'none'
			});
			</script>
			</head>
			<body style="">
			<header style="text-align:center;">
			<a href="../../" style="text-decoration: none;">
			<div id="logo">
			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
			</div>
			</a>
			<p class="author">
			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
			</p>
			</header>

			<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.3.20.$ Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as $6400$ km. The latitude of St. Petersburg is $60^\circ$.
			</p>

			<h3>Solution</h3>
			<p>

			<center>
			<figure>
			<img src="https://savchenkosolutions.com/1/1.3.20/drawing.png" id="drawing"
			loading="lazy" width="250" />
			</figure>
			</center>
			<p>
			The Earth rotates around its axis with a period of $T = 24\;h$. Thus, the cyclic frequency of rotation is
			</p>
			<p class="exp">
			$$ \omega = \frac{2 \pi}{T} = 7.3 \cdot 10^{-5} \; c^{-1} $$
			</p>
			<p>
			At the equator $\varphi = 0^{\circ}$, the distance to the rotation axis is $r =R$. And the speed will be equal to
			</p>
			$$v = \omega R=1675\text{ km/h, }$$
			<p>
			The acceleration associated with daily rotation is found as
		@@ -77,13 +77,13 @@
			</p>
			$$a = \omega^2 R=0{,}034\text{ m/s}^2$$
			<p>
		−	At latitude $\varphi = 60^{\circ}$, the distance to the rotation axis is $r =R\cdot \cos \varphi$. And the speed will be equal to
		+	At latitude $\varphi = 60^{\circ}$, the distance to the rotation axis is $r =R\cdot \cos\varphi$. And the speed will be equal to
			</p>
		−	$$v = \omega R\cdot \cos \varphi=838\text{ km/h, }$$
		+	$$v = \omega R\cdot \cos\varphi=838\text{ km/h, }$$
			<p>
			The acceleration associated with daily rotation will be found as
			</p>
		−	$$a = \omega^2 R\cdot \cos \varphi=0{,}017\text{ m/s}^2$$
		+	$$a = \omega^2 R\cdot \cos\varphi=0{,}017\text{ m/s}^2$$
			</p>

			<h4>Answer</h4>
			<p>
			$$\begin{array}{l}v_\text{e}=1675\text{ km/h, }a_\text{e}=0{,}034\text{ m/s}^2\\v_{\text{l}}=838\text{ km/h, }a_{\text{l}}=0{,}017\text{ m/s}^2\end{array}$$
			</p>


			<footer class="row container">
			<br>
			<p>
			<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
			</p>
			<p>
			<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
			</p>
			</footer>
			</body>

			</html>
		ещё строк без изменений 17

	<!DOCTYPE html>		<!DOCTYPE html>
	<html lang="en">		<html lang="en">

	<head>		<head>
	<meta charset="utf-8">		<meta charset="utf-8">
	<meta name="viewport" content="width=device-width, initial-scale=1.0">		<meta name="viewport" content="width=device-width, initial-scale=1.0">
	<meta http-equiv="content-language" content="en">		<meta http-equiv="content-language" content="en">
	<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">		<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
	<meta name="description" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">		<meta name="description" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">
	<meta name="author" content="Aliaksandr Melnichenka">		<meta name="author" content="Aliaksandr Melnichenka">
	<meta name="date" content="2023-10" scheme="YYYY-MM">		<meta name="date" content="2023-10" scheme="YYYY-MM">
	<meta property="og:title" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">		<meta property="og:title" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">
	<meta property="og:image" content="img/logo.png">		<meta property="og:image" content="img/logo.png">
	<meta property="og:description" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">		<meta property="og:description" content="Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.">
	<meta name="yandex-verification" content="6cfda41f74038368">		<meta name="yandex-verification" content="6cfda41f74038368">
	<title>Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.</title>		<title>Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as 6400 km. The latitude of St. Petersburg is 60^\circ.</title>
	<link rel="stylesheet" href="../../css/css-latex/style.css">		<link rel="stylesheet" href="../../css/css-latex/style.css">
	<link rel="icon" href="../../img/logo.png" type="image/png">		<link rel="icon" href="../../img/logo.png" type="image/png">
	<script src="../../js/jquery-1.10.1.min.js"></script>		<script src="../../js/jquery-1.10.1.min.js"></script>
	<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>		<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
	<script type="text/x-mathjax-config">		<script type="text/x-mathjax-config">
	MathJax.Hub.Config({		MathJax.Hub.Config({
	extensions: ['tex2jax.js'],		extensions: ['tex2jax.js'],
	jax: ['input/TeX', 'output/HTML-CSS'],		jax: ['input/TeX', 'output/HTML-CSS'],
	tex2jax: {		tex2jax: {
	inlineMath: [['$', '$'], ['$', '$']],		inlineMath: [['$', '$'], ['$', '$']],
	processEscapes: true,		processEscapes: true,
	processClass: 'tex2jax',		processClass: 'tex2jax',
	ignoreClass: 'html'		ignoreClass: 'html'
	},		},
	showProcessingMessages: false,		showProcessingMessages: false,
	messageStyle: 'none'		messageStyle: 'none'
	});		});
	</script>		</script>
	</head>		</head>
	<body style="">		<body style="">
	<header style="text-align:center;">		<header style="text-align:center;">
	<a href="../../" style="text-decoration: none;">		<a href="../../" style="text-decoration: none;">
	<div id="logo">		<div id="logo">
	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>		<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
	</div>		</div>
	</a>		</a>
	<p class="author">		<p class="author">
	Solutions of Savchenko Problems in Physics <br>		Solutions of Savchenko Problems in Physics <br>
	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
	</p>		</p>
	</header>		</header>

	<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.3.20.$ Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as $6400$ km. The latitude of St. Petersburg is $60^\circ$.		$1.3.20.$ Determine the speed and acceleration that points on the Earth's surface at the equator and in St. Petersburg due to the Earth's participation in the diurnal rotation. Take the Earth's radius as $6400$ km. The latitude of St. Petersburg is $60^\circ$.
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>

	<center>		<center>
	<figure>		<figure>
	<img src="https://savchenkosolutions.com/1/1.3.20/drawing.png" id="drawing"		<img src="https://savchenkosolutions.com/1/1.3.20/drawing.png" id="drawing"
	loading="lazy" width="250" />		loading="lazy" width="250" />
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	The Earth rotates around its axis with a period of $T = 24\;h$. Thus, the cyclic frequency of rotation is		The Earth rotates around its axis with a period of $T = 24\;h$. Thus, the cyclic frequency of rotation is
	</p>		</p>
	<p class="exp">		<p class="exp">
	$$ \omega = \frac{2 \pi}{T} = 7.3 \cdot 10^{-5} \; c^{-1} $$		$$ \omega = \frac{2 \pi}{T} = 7.3 \cdot 10^{-5} \; c^{-1} $$
	</p>		</p>
	<p>		<p>
	At the equator $\varphi = 0^{\circ}$, the distance to the rotation axis is $r =R$. And the speed will be equal to		At the equator $\varphi = 0^{\circ}$, the distance to the rotation axis is $r =R$. And the speed will be equal to
	</p>		</p>
	$$v = \omega R=1675\text{ km/h, }$$		$$v = \omega R=1675\text{ km/h, }$$
	<p>		<p>
	The acceleration associated with daily rotation is found as		The acceleration associated with daily rotation is found as
	@@ -77,13 +77,13 @@
	</p>		</p>
	$$a = \omega^2 R=0{,}034\text{ m/s}^2$$		$$a = \omega^2 R=0{,}034\text{ m/s}^2$$
	<p>		<p>
	At latitude $\varphi = 60^{\circ}$, the distance to the rotation axis is $r =R\cdot \cos \varphi$. And the speed will be equal to		At latitude $\varphi = 60^{\circ}$, the distance to the rotation axis is $r =R\cdot \cos\varphi$. And the speed will be equal to
	</p>		</p>
	$$v = \omega R\cdot \cos \varphi=838\text{ km/h, }$$		$$v = \omega R\cdot \cos\varphi=838\text{ km/h, }$$
	<p>		<p>
	The acceleration associated with daily rotation will be found as		The acceleration associated with daily rotation will be found as
	</p>		</p>
	$$a = \omega^2 R\cdot \cos \varphi=0{,}017\text{ m/s}^2$$		$$a = \omega^2 R\cdot \cos\varphi=0{,}017\text{ m/s}^2$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$\begin{array}{l}v_\text{e}=1675\text{ km/h, }a_\text{e}=0{,}034\text{ m/s}^2\\v_{\text{l}}=838\text{ km/h, }a_{\text{l}}=0{,}017\text{ m/s}^2\end{array}$$		$$\begin{array}{l}v_\text{e}=1675\text{ km/h, }a_\text{e}=0{,}034\text{ m/s}^2\\v_{\text{l}}=838\text{ km/h, }a_{\text{l}}=0{,}017\text{ m/s}^2\end{array}$$
	</p>		</p>


	<footer class="row container">		<footer class="row container">
	<br>		<br>
	<p>		<p>
	<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>		<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
	</p>		</p>
	<p>		<p>
	<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>		<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
	</p>		</p>
	</footer>		</footer>
	</body>		</body>

	</html>		</html>
	ещё строк без изменений 17