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<item xml:lang="es">
		<title>Force a man must exert to pull a sled (8334)</title>
		<link>https://www.ejercicios-fyq.com/Force-a-man-must-exert-to-pull-a-sled-8334</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Force-a-man-must-exert-to-pull-a-sled-8334</guid>
		<dc:date>2024-10-23T03:01:15Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Dynamics</dc:subject>
		<dc:subject>SOLVED</dc:subject>
		<dc:subject>Friction force</dc:subject>
		<dc:subject>Newton's second law</dc:subject>

		<description>
&lt;p&gt;A man pulls a sled up a ramp using a rope attached to the front, as illustrated in the figure. &lt;br class='autobr' /&gt;
The sled has a mass of 80 kg. The kinetic friction coefficient between the sled and the ramp is (), the angle between the ramp and the horizontal is , and the angle between the rope and the ramp is . What force must the man exert to keep the sled moving at a constant speed?&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A man pulls a sled up a ramp using a rope attached to the front, as illustrated in the figure.&lt;/p&gt;
&lt;div class='spip_document_1348 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L295xH258/zzz-790a0.jpg?1758420408' width='295' height='258' alt='' /&gt;
&lt;/figure&gt;
&lt;/div&gt;
&lt;p&gt;The sled has a mass of 80 kg. The kinetic friction coefficient between the sled and the ramp is (&lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L68xH16/64cc50cc8344bf421cebea37a6b3ed33-809ba.png?1732956198' style='vertical-align:middle;' width='68' height='16' alt=&#034;\mu_k = 0.70&#034; title=&#034;\mu_k = 0.70&#034; /&gt;), the angle between the ramp and the horizontal is &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L22xH13/f13e28ce829b262174de9bc35f8a8b7d-c9043.png?1732956198' style='vertical-align:middle;' width='22' height='13' alt=&#034;25^o&#034; title=&#034;25^o&#034; /&gt;, and the angle between the rope and the ramp is &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L32xH42/9a852d58dca44420883431ed11b2f8b2-559a4.png?1732956198' style='vertical-align:middle;' width='32' height='42' alt=&#034;35^o&#034; title=&#034;35^o&#034; /&gt;. What force must the man exert to keep the sled moving at a constant speed?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;To analyze the forces acting on the sled, decompose them according to the system formed by the direction of movement and a perpendicular axis: &lt;br/&gt;&lt;/p&gt;
&lt;div class='spip_document_1349 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;img src='https://www.ejercicios-fyq.com/IMG/jpg/z.jpg' width=&#034;268&#034; height=&#034;228&#034; alt='' /&gt;
&lt;/figure&gt;
&lt;/div&gt; &lt;p&gt;&lt;br/&gt; The components of the weight and applied force are: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/914aec00087a7a6616bff968b6c05f15.png' style=&#034;vertical-align:middle;&#034; width=&#034;195&#034; height=&#034;52&#034; alt=&#034;\left p_x = m\cdot g\cdot sen\ 25 ^o \atop p_y = m\cdot g\cdot cos\ 25^o \right \}&#034; title=&#034;\left p_x = m\cdot g\cdot sen\ 25 ^o \atop p_y = m\cdot g\cdot cos\ 25^o \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/91d1053de8a842a0a58213a1e1f66b56.png' style=&#034;vertical-align:middle;&#034; width=&#034;167&#034; height=&#034;52&#034; alt=&#034;\left F_x = F\cdot cos\ 35 ^o \atop F_y = F\cdot sen\ 35^o \right \}&#034; title=&#034;\left F_x = F\cdot cos\ 35 ^o \atop F_y = F\cdot sen\ 35^o \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; In the direction perpendicular to the ramp, the sum of the forces must be zero. Consider the &#034;y-component&#034; of the weight to solve for the normal force: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/5d6c3b0c3fa28a8ac5a18a07da98c6a9.png' style=&#034;vertical-align:middle;&#034; width=&#034;481&#034; height=&#034;22&#034; alt=&#034;p_y = N + F_y\ \to\ \color[RGB]{2,112,20}{\bf N = m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o}}&#034; title=&#034;p_y = N + F_y\ \to\ \color[RGB]{2,112,20}{\bf N = m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; In the direction of movement, the sum of the forces must also be zero because the sled moves at a constant speed. This gives: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/d08ce8c4ff274e375a8a4dbf748ed443.png' style=&#034;vertical-align:middle;&#034; width=&#034;347&#034; height=&#034;20&#034; alt=&#034;F_x - p_x- F_R = 0\ \to\ \color[RGB]{2,112,20}{\bm{F_x= p_x - F_R}}&#034; title=&#034;F_x - p_x- F_R = 0\ \to\ \color[RGB]{2,112,20}{\bm{F_x= p_x - F_R}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The friction force is the product of the normal force and the coefficient of friction: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/a461c7975b0794fae49e6685e5a73b09.png' style=&#034;vertical-align:middle;&#034; width=&#034;571&#034; height=&#034;23&#034; alt=&#034;F\cdot cos\ 35^o = \mu_c(m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o) + m\cdot g\cdot sen\ 25^o&#034; title=&#034;F\cdot cos\ 35^o = \mu_c(m\cdot g\cdot cos\ 25^o - F\cdot sen\ 35^o) + m\cdot g\cdot sen\ 25^o&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substitute the known values into the equations to solve for the force exerted by the man: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/042a3cabf1ae528823e6b293e081baf7.png' style=&#034;vertical-align:middle;&#034; width=&#034;680&#034; height=&#034;28&#034; alt=&#034;0.819F = 497.4 - 0.402F + 331.3\ \to\ 1.221F = 828.7\ \to\ \fbox{\color[RGB]{192,0,0}{\bf F= 678.7\ N}}&#034; title=&#034;0.819F = 497.4 - 0.402F + 331.3\ \to\ 1.221F = 828.7\ \to\ \fbox{\color[RGB]{192,0,0}{\bf F= 678.7\ N}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Acceleration and time taken by a car to increase speed (8329)</title>
		<link>https://www.ejercicios-fyq.com/Acceleration-and-time-taken-by-a-car-to-increase-speed-8329</link>
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		<dc:date>2024-10-10T03:34:15Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MRUA</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>Acceleration</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A car, which has uniformly accelerated motion, increases its speed from 18 km/h to 72 km/h over a straight distance of 37.5 meters. Calculate the time taken for this journey and its acceleration.&lt;/p&gt;


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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A car, which has uniformly accelerated motion, increases its speed from 18 km/h to 72 km/h over a straight distance of 37.5 meters. Calculate the time taken for this journey and its acceleration.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;To make the problem homogeneous, the first step is to express the speeds in SI units: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/67c9d605de0bbcc641dfbf99e22bae3c.png' style=&#034;vertical-align:middle;&#034; width=&#034;398&#034; height=&#034;105&#034; alt=&#034;\left 18\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{5\ m\cdot s^{-1}}}} \atop 72\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{20\ m\cdot s^{-1}}}} \right \}&#034; title=&#034;\left 18\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{5\ m\cdot s^{-1}}}} \atop 72\ \dfrac{\cancel{km}}{\cancel{h}}\cdot \dfrac{10^3\ m}{1\ \cancel{km}}\cdot \dfrac{1\ \cancel{h}}{3.6\cdot 10^3\ s} = {\color[RGB]{0,112,192}{\bm{20\ m\cdot s^{-1}}}} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; You can relate the change in speed and the distance covered with the car's acceleration: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/a55b922add5dd0650cc307c0d373c97c.png' style=&#034;vertical-align:middle;&#034; width=&#034;317&#034; height=&#034;53&#034; alt=&#034;v_f^2 = v_i^2 + 2ad\ \to\ \color[RGB]{2,112,20}{\bm{a = \frac{(v_f^2 - v_i^2)}{2d}}}&#034; title=&#034;v_f^2 = v_i^2 + 2ad\ \to\ \color[RGB]{2,112,20}{\bm{a = \frac{(v_f^2 - v_i^2)}{2d}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substitute the values and calculate: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/c1facad935205b9ca3ac232a004e0209.png' style=&#034;vertical-align:middle;&#034; width=&#034;350&#034; height=&#034;51&#034; alt=&#034;a = \frac{(20^2 - 5^2)\ m\cancel{^2}\cdot s^{-2}}{2\cdot 37.5\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{5\ m\cdot s^{-2}}}}&#034; title=&#034;a = \frac{(20^2 - 5^2)\ m\cancel{^2}\cdot s^{-2}}{2\cdot 37.5\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{5\ m\cdot s^{-2}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The time needed to make this speed change is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/14fb46f95265619345e66352a4f43496.png' style=&#034;vertical-align:middle;&#034; width=&#034;289&#034; height=&#034;44&#034; alt=&#034;v_f = v_i + a\cdot t\ \to\ \color[RGB]{2,112,20}{\bm{t = \frac{v_f - v_i}{a}}}&#034; title=&#034;v_f = v_i + a\cdot t\ \to\ \color[RGB]{2,112,20}{\bm{t = \frac{v_f - v_i}{a}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substitute and calculate: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/deca9ca7581aa024e04fec4a7e96bc1f.png' style=&#034;vertical-align:middle;&#034; width=&#034;257&#034; height=&#034;48&#034; alt=&#034;t = \frac{(20 - 5)\ \cancel{m}\cdot \cancel{s^{-1}}}{5\ m\cdot s^{\cancel{-2}}} = \fbox{\color[RGB]{192,0,0}{\bf 3\ s}}&#034; title=&#034;t = \frac{(20 - 5)\ \cancel{m}\cdot \cancel{s^{-1}}}{5\ m\cdot s^{\cancel{-2}}} = \fbox{\color[RGB]{192,0,0}{\bf 3\ s}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Kinetic energy of a moving projectile (8317)</title>
		<link>https://www.ejercicios-fyq.com/Kinetic-energy-of-a-moving-projectile-8317</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Kinetic-energy-of-a-moving-projectile-8317</guid>
		<dc:date>2024-09-19T04:20:43Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Kinetic energy</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;Calculate the kinetic energy of an 18 g projectile moving horizontally at .&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calculate the kinetic energy of an 18 g projectile moving horizontally at &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L90xH20/0292e51078171553d6decc64fbd4fd45-2cc57.png?1732956060' style='vertical-align:middle;' width='90' height='20' alt=&#034;97\ m\cdot s^{-1}&#034; title=&#034;97\ m\cdot s^{-1}&#034; /&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;If you observe the units in the problem statement, you will see that the mass is not expressed in the SI unit of mass. The first thing you need to do is convert it to &lt;i&gt;kg&lt;/i&gt; to make the problem homogeneous. &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/7d935963158a5faf7edf1f70520d19b8.png' style=&#034;vertical-align:middle;&#034; width=&#034;219&#034; height=&#034;51&#034; alt=&#034;18\ \cancel{g}\cdot \frac{1\ kg}{10^3\ \cancel{g}} = \color[RGB]{0,112,192}{\bf 0.018\ kg}&#034; title=&#034;18\ \cancel{g}\cdot \frac{1\ kg}{10^3\ \cancel{g}} = \color[RGB]{0,112,192}{\bf 0.018\ kg}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The kinetic energy of the projectile is given by the equation: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/33d2c0c292dc2a67768a18f873cdc4af.png' style=&#034;vertical-align:middle;&#034; width=&#034;122&#034; height=&#034;43&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{E_k= \frac{m}{2}\cdot v^2}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{E_k= \frac{m}{2}\cdot v^2}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; You know the necessary data, so you just need to substitute and calculate: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/be5978559ab337f32b73f5bf8b014077.png' style=&#034;vertical-align:middle;&#034; width=&#034;366&#034; height=&#034;44&#034; alt=&#034;E_k = \frac{0.018\ kg}{2}\cdot 97^2\ m^2\cdot s^{-2} = \fbox{\color[RGB]{192,0,0}{\bf 84.7\ J}}&#034; title=&#034;E_k = \frac{0.018\ kg}{2}\cdot 97^2\ m^2\cdot s^{-2} = \fbox{\color[RGB]{192,0,0}{\bf 84.7\ J}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Calculation of braking acceleration (8312)</title>
		<link>https://www.ejercicios-fyq.com/Calculation-of-braking-acceleration-8312</link>
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		<dc:date>2024-09-16T03:46:33Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MRUA</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>Acceleration</dc:subject>
		<dc:subject>Uniformly accelerated rectilinear motion</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;Determine the acceleration of a car, initially moving at a speed of 120 km/h, knowing that it takes 20 seconds to come to a complete stop.&lt;/p&gt;


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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Determine the acceleration of a car, initially moving at a speed of 120 km/h, knowing that it takes 20 seconds to come to a complete stop.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;The car will have changed its speed by 120 km/h in those 20 seconds. First, convert the speed to International System units: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/3c92d2cfff9f9d1e10ef3bf400fdcb48.png' style=&#034;vertical-align:middle;&#034; width=&#034;356&#034; height=&#034;50&#034; alt=&#034;120\ \frac{\cancel{km}}{\cancel{h}}\cdot \frac{1\ 000\ m}{1\ \cancel{km}}\cdot \frac{1\ \cancel{h}}{3\ 600\ s} = \color[RGB]{0,112,192}{\bm{33.3\ \frac{m}{s}}}&#034; title=&#034;120\ \frac{\cancel{km}}{\cancel{h}}\cdot \frac{1\ 000\ m}{1\ \cancel{km}}\cdot \frac{1\ \cancel{h}}{3\ 600\ s} = \color[RGB]{0,112,192}{\bm{33.3\ \frac{m}{s}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The acceleration will be: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/bedcd86f3117d8395e13cd95675ccf3d.png' style=&#034;vertical-align:middle;&#034; width=&#034;353&#034; height=&#034;49&#034; alt=&#034;a = \frac{\Delta v}{\Delta t} = \frac{(0 - 33.3)\ \frac{m}{s}}{20\ s} = \fbox{\color[RGB]{192,0,0}{\bm{-1.67\ \frac{m}{s^2}}}}&#034; title=&#034;a = \frac{\Delta v}{\Delta t} = \frac{(0 - 33.3)\ \frac{m}{s}}{20\ s} = \fbox{\color[RGB]{192,0,0}{\bm{-1.67\ \frac{m}{s^2}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Principle of dimensional homogeneity (8304)</title>
		<link>https://www.ejercicios-fyq.com/Principle-of-dimensional-homogeneity-8304</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Principle-of-dimensional-homogeneity-8304</guid>
		<dc:date>2024-09-11T10:46:47Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Dimensions</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;What is the principle of dimensional homogeneity?&lt;/p&gt;


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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;What is the &lt;i&gt;principle of dimensional homogeneity&lt;/i&gt;?&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;This principle states that equations relating physical quantities must be consistent, meaning that the same dimensions must appear on both sides of the equation. The best way to understand this is through an example. &lt;br/&gt; &lt;br/&gt; The speed at which an object moves under the influence of acceleration is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/e0b460db1a312f718d70061752740edb.png' style=&#034;vertical-align:middle;&#034; width=&#034;117&#034; height=&#034;19&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{v = v_0 + at}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{v = v_0 + at}}&#034; /&gt; (Ec. 1) &lt;br/&gt; &lt;br/&gt; Speed is the ratio of distance to the time taken to cover it. In terms of dimensions, it is expressed as: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/ea54684dba160033962f928c8fcfaee1.png' style=&#034;vertical-align:middle;&#034; width=&#034;88&#034; height=&#034;55&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{[v] = \frac{[L]}{[t]}}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{[v] = \frac{[L]}{[t]}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Acceleration is the ratio of the change in speed to the time taken for that change, so it can be expressed as: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/c77f5d751b5213f8ec8fc9dc4121f887.png' style=&#034;vertical-align:middle;&#034; width=&#034;91&#034; height=&#034;55&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{[a] = \frac{[L]}{[t]^2}}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{[a] = \frac{[L]}{[t]^2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; If you apply this to equation (Eq. 1), you will have: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/6b3f5fdd7ad7d86f719f461d86c8bc2e.png' style=&#034;vertical-align:middle;&#034; width=&#034;388&#034; height=&#034;55&#034; alt=&#034;\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]^2}\cdot [t]\ \to\ \color[RGB]{192,0,0}{\bm{\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]}}}&#034; title=&#034;\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]^2}\cdot [t]\ \to\ \color[RGB]{192,0,0}{\bm{\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; As you can see, the dimensions are the same on both sides of the equation.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Efficiency and useful power (8300)</title>
		<link>https://www.ejercicios-fyq.com/Efficiency-and-useful-power-8300</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Efficiency-and-useful-power-8300</guid>
		<dc:date>2024-09-07T03:10:02Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Power</dc:subject>
		<dc:subject>Efficiency</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A motor is supplied with a power of 1 000 W. If the efficiency of this motor is , calculate the useful power and the lost power.&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/Energy" rel="directory"&gt;Energy&lt;/a&gt;

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&lt;a href="https://www.ejercicios-fyq.com/Power" rel="tag"&gt;Power&lt;/a&gt;, 
&lt;a href="https://www.ejercicios-fyq.com/Efficiency" rel="tag"&gt;Efficiency&lt;/a&gt;, 
&lt;a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A motor is supplied with a power of 1 000 W. If the efficiency of this motor is &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L32xH14/e3de74591a3bb01cc6057dd4cfc8eff5-34645.png?1732958364' style='vertical-align:middle;' width='32' height='14' alt=&#034;80\ \%&#034; title=&#034;80\ \%&#034; /&gt;, calculate the useful power and the lost power.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Efficiency or performance is the ratio between the useful power and the supplied power: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/916d754a2a89d8305f376635540b523b.png' style=&#034;vertical-align:middle;&#034; width=&#034;130&#034; height=&#034;51&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\rho = \frac{P_u}{P_T}\cdot 100}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\rho = \frac{P_u}{P_T}\cdot 100}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The useful power will be: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/bf41b7fbd9e3f8f16ea4a6e5c71bf7cf.png' style=&#034;vertical-align:middle;&#034; width=&#034;389&#034; height=&#034;45&#034; alt=&#034;P_u = \frac{80}{100}\cdot P_T = 0.8\cdot 1\ 000\ W = \fbox{\color[RGB]{192,0,0}{\bf 800\ W}}&#034; title=&#034;P_u = \frac{80}{100}\cdot P_T = 0.8\cdot 1\ 000\ W = \fbox{\color[RGB]{192,0,0}{\bf 800\ W}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The lost power will be the difference between the supplied and the used power: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/fe923c3b43cea3c5a7cf254d822c035f.png' style=&#034;vertical-align:middle;&#034; width=&#034;262&#034; height=&#034;28&#034; alt=&#034;(1\ 000 - 800)\ W = \fbox{\color[RGB]{192,0,0}{\bf 200\ W}}&#034; title=&#034;(1\ 000 - 800)\ W = \fbox{\color[RGB]{192,0,0}{\bf 200\ W}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Calculation of the half-life of a radioactive element (8299)</title>
		<link>https://www.ejercicios-fyq.com/Calculation-of-the-half-life-of-a-radioactive-element-8299</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Calculation-of-the-half-life-of-a-radioactive-element-8299</guid>
		<dc:date>2024-09-04T03:49:25Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Radioactive activity</dc:subject>
		<dc:subject>Nuclear physics</dc:subject>
		<dc:subject>Half-life period</dc:subject>
		<dc:subject>Radioactive</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A sample of 300 grams of a radioactive element remains 18.75 grams after 24 hours. Calculate the half-life.&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/Nuclear-physics-282" rel="directory"&gt;Nuclear physics&lt;/a&gt;

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&lt;a href="https://www.ejercicios-fyq.com/Half-life-period" rel="tag"&gt;Half-life period&lt;/a&gt;, 
&lt;a href="https://www.ejercicios-fyq.com/Radioactive" rel="tag"&gt;Radioactive&lt;/a&gt;, 
&lt;a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A sample of 300 grams of a radioactive element remains 18.75 grams after 24 hours. Calculate the half-life.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;You must use the decay law, but referring to the mass of the substance: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/a7837110049bd6922afc8fc688568fce.png' style=&#034;vertical-align:middle;&#034; width=&#034;149&#034; height=&#034;23&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{m = m_0\cdot e^{- \lambda\cdot t}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{m = m_0\cdot e^{- \lambda\cdot t}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Solving for the value of &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/c6a6eb61fd9c6c913da73b3642ca147d.png' style=&#034;vertical-align:middle;&#034; width=&#034;18&#034; height=&#034;40&#034; alt=&#034;\lambda&#034; title=&#034;\lambda&#034; /&gt;, you get: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/d4cf9a033527abef837296344e1fafe8.png' style=&#034;vertical-align:middle;&#034; width=&#034;302&#034; height=&#034;57&#034; alt=&#034;- \lambda = \frac{ln \frac{18.75\ \cancel{g}}{300\ \cancel{g}}}{86\ 400\ s} = \color[RGB]{0,112,192}{\bm{3.2\cdot 10^{-5}\ s^{-1}}}&#034; title=&#034;- \lambda = \frac{ln \frac{18.75\ \cancel{g}}{300\ \cancel{g}}}{86\ 400\ s} = \color[RGB]{0,112,192}{\bm{3.2\cdot 10^{-5}\ s^{-1}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; You know that radioactive activity is related to the half-life and this to the half-life period: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/1370bc65dcc71d914e193ccf3b8e984b.png' style=&#034;vertical-align:middle;&#034; width=&#034;90&#034; height=&#034;65&#034; alt=&#034;\left \lambda = \frac{1}{\tau} \atop \tau = \frac{t_{1/2}}{ln\ 2} \right \}&#034; title=&#034;\left \lambda = \frac{1}{\tau} \atop \tau = \frac{t_{1/2}}{ln\ 2} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Substituting and solving: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/711305399e214577f0e8adc0477070ef.png' style=&#034;vertical-align:middle;&#034; width=&#034;229&#034; height=&#034;45&#034; alt=&#034;t_{1/2} = \frac{ln\ 2}{\lambda} = \fbox{\color[RGB]{192,0,0}{\bf 21\ 661\ s}}&#034; title=&#034;t_{1/2} = \frac{ln\ 2}{\lambda} = \fbox{\color[RGB]{192,0,0}{\bf 21\ 661\ s}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Expressed in hours, it will be: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/ab3616446fa7cd3f753a71f3383d802a.png' style=&#034;vertical-align:middle;&#034; width=&#034;299&#034; height=&#034;45&#034; alt=&#034;t_{1/2} = 21\ 661\ \cancel{s}\cdot \frac{1\ h}{3\ 600\ s} = \fbox{\color[RGB]{192,0,0}{\bf 6\ h}}&#034; title=&#034;t_{1/2} = 21\ 661\ \cancel{s}\cdot \frac{1\ h}{3\ 600\ s} = \fbox{\color[RGB]{192,0,0}{\bf 6\ h}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Vertical upward lunch (8289)</title>
		<link>https://www.ejercicios-fyq.com/Vertical-upward-lunch-8289</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Vertical-upward-lunch-8289</guid>
		<dc:date>2024-08-15T05:47:50Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MRUA</dc:subject>
		<dc:subject>Kinematics</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A body is launched vertically upwards with an initial velocity of . How long will it take to reach its maximum height?&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/Movements" rel="directory"&gt;Movements&lt;/a&gt;

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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A body is launched vertically upwards with an initial velocity of &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L90xH20/7885d5c2ff711290f70bf3c1f851d928-95980.png?1733114143' style='vertical-align:middle;' width='90' height='20' alt=&#034;70\ m\cdot s^{-1}&#034; title=&#034;70\ m\cdot s^{-1}&#034; /&gt;. How long will it take to reach its maximum height?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Since it is a vertical upward launch, if you take the initial velocity as positive, the gravitational acceleration must be negative. The velocity of the object at any instant can be obtained from the expression: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/166981455fdbad406161aa3e5ccd8b5f.png' style=&#034;vertical-align:middle;&#034; width=&#034;116&#034; height=&#034;19&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{v = v_0 - gt}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{v = v_0 - gt}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The condition for the body to stop ascending is that the velocity is zero; at that moment, it will have reached its maximum height. By imposing this condition on the previous equation, you can solve for the time and calculate it: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/0f021d446b709ce49eb299ad95040992.png' style=&#034;vertical-align:middle;&#034; width=&#034;383&#034; height=&#034;70&#034; alt=&#034;g\cdot t = v_0\ \to\ {\color[RGB]{2,112,20}{\bm{t = \frac{v_0}{g}}}} = \frac{70\ \frac{\cancel{m}}{\cancel{s}}}{9.8\ \frac{\cancel{m}}{s\cancel{^2}}}= \fbox{\color[RGB]{192,0,0}{\bf 7.14\ s}}&#034; title=&#034;g\cdot t = v_0\ \to\ {\color[RGB]{2,112,20}{\bm{t = \frac{v_0}{g}}}} = \frac{70\ \frac{\cancel{m}}{\cancel{s}}}{9.8\ \frac{\cancel{m}}{s\cancel{^2}}}= \fbox{\color[RGB]{192,0,0}{\bf 7.14\ s}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Application of Torricelli's theorem (8284)</title>
		<link>https://www.ejercicios-fyq.com/Application-of-Torricelli-s-theorem-8284</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Application-of-Torricelli-s-theorem-8284</guid>
		<dc:date>2024-08-12T02:53:41Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Torricelli's theorem</dc:subject>
		<dc:subject>Horizontal launch</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A cylindrical container is filled with a liquid up to a height of one meter from the base of the container. Then, a hole is made at a point 80 cm below the liquid level: &lt;br class='autobr' /&gt;
a) What is the exit velocity of the liquid through the hole? &lt;br class='autobr' /&gt;
b) At what distance from the container will the first drop of liquid that touches the ground fall?&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A cylindrical container is filled with a liquid up to a height of one meter from the base of the container. Then, a hole is made at a point 80 cm below the liquid level:&lt;/p&gt;
&lt;p&gt;a) What is the exit velocity of the liquid through the hole?&lt;/p&gt;
&lt;p&gt;b) At what distance from the container will the first drop of liquid that touches the ground fall?&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) The exit velocity of the liquid through the hole is given by the expression: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/6b8ef53b9924732e6103f56f9f68c5d6.png' style=&#034;vertical-align:middle;&#034; width=&#034;141&#034; height=&#034;26&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{v= \sqrt{2\cdot g\cdot h}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{v= \sqrt{2\cdot g\cdot h}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Since the hole is made at a distance of 0.8 m from the liquid level: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/4255c299f7ff628166fc0ce03791540a.png' style=&#034;vertical-align:middle;&#034; width=&#034;324&#034; height=&#034;52&#034; alt=&#034;v = \sqrt{2\cdot 9.8\ \frac{m}{s^2}\cdot 0.8\ m} = \fbox{\color[RGB]{192,0,0}{\bm{3.96\ \frac{m}{s}}}}&#034; title=&#034;v = \sqrt{2\cdot 9.8\ \frac{m}{s^2}\cdot 0.8\ m} = \fbox{\color[RGB]{192,0,0}{\bm{3.96\ \frac{m}{s}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) To calculate the distance at which the first drop falls, you must consider that it follows a motion similar to a horizontal launch. In this case, the position with respect to the X and Y axes follows the equations: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/031243fec7947d73e589855882668bd5.png' style=&#034;vertical-align:middle;&#034; width=&#034;105&#034; height=&#034;54&#034; alt=&#034;\left {\color[RGB]{2,112,20}{\bf x = v\cdot t}} \atop {\color[RGB]{2,112,20}{\bm{y = \frac{1}{2}gt^2}}}} \right \}&#034; title=&#034;\left {\color[RGB]{2,112,20}{\bf x = v\cdot t}} \atop {\color[RGB]{2,112,20}{\bm{y = \frac{1}{2}gt^2}}}} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Since you know that the drop starts at a height of 0.2 m: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/8cc8244ea60608b15a7b13fca5849665.png' style=&#034;vertical-align:middle;&#034; width=&#034;291&#034; height=&#034;65&#034; alt=&#034;t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2\cdot 0.2\ \cancel{m}}{9.8\ \frac{\cancel{m}}{s^2}}}= \color[RGB]{0,112,192}{\bf 0.2\ s}&#034; title=&#034;t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2\cdot 0.2\ \cancel{m}}{9.8\ \frac{\cancel{m}}{s^2}}}= \color[RGB]{0,112,192}{\bf 0.2\ s}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; To find the horizontal position, substitute this time: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/43a81991bb4b60becdd55b81eb5a1c8b.png' style=&#034;vertical-align:middle;&#034; width=&#034;343&#034; height=&#034;45&#034; alt=&#034;x = v\cdot t = 3.96\ \frac{m}{\cancel{s}}\cdot 0.2\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bf 0.79\ m}}&#034; title=&#034;x = v\cdot t = 3.96\ \frac{m}{\cancel{s}}\cdot 0.2\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bf 0.79\ m}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>Relationship between linear and angular quantities (7425)</title>
		<link>https://www.ejercicios-fyq.com/Relationship-between-linear-and-angular-quantities-7425</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Relationship-between-linear-and-angular-quantities-7425</guid>
		<dc:date>2021-12-10T07:07:49Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>MRUA</dc:subject>
		<dc:subject>MCUA</dc:subject>
		<dc:subject>EDICO</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A car starts at and uniformly accelerates at for 5 s. &lt;br class='autobr' /&gt;
a) Calculate the final velocity and the displacement of the car. &lt;br class='autobr' /&gt;
b) Determine the diameter of the tire if the angular displacement is 60 radians. &lt;br class='autobr' /&gt;
c) Calculate the final angular velocity and angular acceleration of the tires.&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A car starts at &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L78xH20/911c7f12083f2f5b5198c5b88e2c87bd-4743a.png?1733073348' style='vertical-align:middle;' width='78' height='20' alt=&#034;1\ m\cdot s^{-1}&#034; title=&#034;1\ m\cdot s^{-1}&#034; /&gt; and uniformly accelerates at &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L96xH20/af5dbc5a2eea86d1b82a58ae99f44962-13acf.png?1733007963' style='vertical-align:middle;' width='96' height='20' alt=&#034;2.5\ m\cdot s^{-2}&#034; title=&#034;2.5\ m\cdot s^{-2}&#034; /&gt; for 5 s.&lt;/p&gt;
&lt;p&gt;a) Calculate the final velocity and the displacement of the car.&lt;/p&gt;
&lt;p&gt;b) Determine the diameter of the tire if the angular displacement is 60 radians.&lt;/p&gt;
&lt;p&gt;c) Calculate the final angular velocity and angular acceleration of the tires.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) The equation for calculating the final velocity is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/32bc1749368305b3e3e85277fba5af75.png' style=&#034;vertical-align:middle;&#034; width=&#034;134&#034; height=&#034;19&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{v= v_0 + a\cdot t}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{v= v_0 + a\cdot t}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Just replace the data and calculate: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/32f16e66a7b2b23bdd6f69d406681458.png' style=&#034;vertical-align:middle;&#034; width=&#034;298&#034; height=&#034;44&#034; alt=&#034;v= 1\ \frac{m}{s} + 2.5\ \frac{m}{s\cancel{^2}}\cdot 5\ \cancel{s} = \fbox{\color[RGB]{192,0,0}{\bm{14\ \frac{m}{s}}}}&#034; title=&#034;v= 1\ \frac{m}{s} + 2.5\ \frac{m}{s\cancel{^2}}\cdot 5\ \cancel{s} = \fbox{\color[RGB]{192,0,0}{\bm{14\ \frac{m}{s}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The equation for calculating the displacement is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/aab59a717a99d28719a64a6ba69f66d4.png' style=&#034;vertical-align:middle;&#034; width=&#034;170&#034; height=&#034;43&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{d= v_0\cdot t + \frac{a}{2}\cdot t^2}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{d= v_0\cdot t + \frac{a}{2}\cdot t^2}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Using the data in the problem statement: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/40d10dde567195268cfdf4dad4ae2a6d.png' style=&#034;vertical-align:middle;&#034; width=&#034;364&#034; height=&#034;50&#034; alt=&#034;d= 1\ \frac{m}{\cancel{s}}\cdot 5\ \cancel{s} + \frac{2.5}{2}\ \frac{m}{\cancel{s^2}}\cdot 5^2\ \cancel{s^2} = \fbox{\color[RGB]{192,0,0}{\bf 36\ m}}&#034; title=&#034;d= 1\ \frac{m}{\cancel{s}}\cdot 5\ \cancel{s} + \frac{2.5}{2}\ \frac{m}{\cancel{s^2}}\cdot 5^2\ \cancel{s^2} = \fbox{\color[RGB]{192,0,0}{\bf 36\ m}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) You must relate the distance covered to the radius of the wheel: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/ee69a671765cfeddd0b72e12907af2e5.png' style=&#034;vertical-align:middle;&#034; width=&#034;360&#034; height=&#034;49&#034; alt=&#034;d= \varphi\cdot R\ \to\ R = \frac{d}{\varphi} = \frac{36\ m}{60} = \color[RGB]{0,112,192}{\bf 0.6\ m}&#034; title=&#034;d= \varphi\cdot R\ \to\ R = \frac{d}{\varphi} = \frac{36\ m}{60} = \color[RGB]{0,112,192}{\bf 0.6\ m}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Diameter is two times the radius: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/75215b9ef5bbbc0c16a98726f667802f.png' style=&#034;vertical-align:middle;&#034; width=&#034;280&#034; height=&#034;27&#034; alt=&#034;D= 2R = 2\cdot 0.6\ m = \fbox{\color[RGB]{192,0,0}{\bf 1.2\ m}}&#034; title=&#034;D= 2R = 2\cdot 0.6\ m = \fbox{\color[RGB]{192,0,0}{\bf 1.2\ m}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) The final angular velocity is: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/913cf07ea0899fa18d7833250496b91c.png' style=&#034;vertical-align:middle;&#034; width=&#034;388&#034; height=&#034;52&#034; alt=&#034;v = \omega\cdot R\ \to\ \omega =\frac{v}{R} = \frac{14\ \frac{\cancel{m}}{s}}{0.6\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{23\ s^{-1}}}}&#034; title=&#034;v = \omega\cdot R\ \to\ \omega =\frac{v}{R} = \frac{14\ \frac{\cancel{m}}{s}}{0.6\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{23\ s^{-1}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The angular acceleration is: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/ca3eb66e459803142a99a2d77c73ce5d.png' style=&#034;vertical-align:middle;&#034; width=&#034;395&#034; height=&#034;52&#034; alt=&#034;a = \alpha\cdot R\ \to\ \alpha = \frac{a}{R}= \frac{2.5\ \frac{\cancel{m}}{s^2}}{0.6\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{4.2\ s^{-2}}}}&#034; title=&#034;a = \alpha\cdot R\ \to\ \alpha = \frac{a}{R}= \frac{2.5\ \frac{\cancel{m}}{s^2}}{0.6\ \cancel{m}} = \fbox{\color[RGB]{192,0,0}{\bm{4.2\ s^{-2}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Descarga el enunciado y la resoluci&#243;n del problema en formato EDICO si lo necesitas&lt;/b&gt;.&lt;/p&gt;
&lt;div class='spip_document_1603 spip_document spip_documents spip_document_file spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt;
&lt;a href=&#034;https://ejercicios-fyq.com/apuntes/descarga.php?file=Ej_7425.edi&#034; class=&#034; spip_doc_lien&#034; title='Zip - ' type=&#034;application/zip&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/plugins-dist/medias/prive/vignettes/zip.svg?1772792240' width='64' height='64' alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/div&gt;
		
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