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your online Student Tools Premium Practice for AP Excellence. Theres a huge collection of challenging questions on the ALBERT website which are completely updated to reflect the new AP Physics 1 curriculum. Both the force $\vec{F}$ and the rode lie in the plane of the page. Correspondingly, the force that the mass $m_2$ exerts on $m_1$ has the same magnitude but in the opposite direction which is down. A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. A The force would remain the same. What is the net torque on the wheel due to these three forces about the axle through $O$ perpendicular to the page? Sign in . Refer to the pdf version to find the explanation. Problem (14): A 2-kg crate is pulled over a rough horizontal surface by the force of $25\,{\rm N}$ which makes an angle of $37^\circ$ with the horizontal. \begin{align*} \tau&=r_{\bot}F \\ &=(L\sin\theta) F \\ &=(4\sin 30^\circ)(10) \\&=20\quad\rm m.N \end{align*}, (d) In this configuration, the angle between the force line and the direction of the rod is $\theta=60^\circ$. The change in the momentum is defined as $\Delta \vec{P}=m(\vec{v}_2-\vec{v}_1)$. From the moment of leaving the cloud to reaching the ground, how does the air resistance force change? This site provides class notes, review sheets, PDF notes and lecture notes. The force would decrease by a factor of 4 4. Thus, in this case, it is better to use the following kinematics equation. Students should be able to analyze situations in which a particle remains at rest, or moves with constant velocity, under the influence of several forces. The companion website for Physics: Principles with Applications by Giancoli. This is an extensive unit. Thus, the $\vec{N}_{12}=-\vec{N}_{21}$. Solution: One of the most common problems on circular motion and gravitation in the AP Physics 1 exam is about whirling a satellite around a planet. Meeting Point- PREDICTION CHALLENGE.doc, 4. Use g = 10 m/s. Now that the mass is known, use the weight formula to find the object's weight on the Moon \begin{align*} W_{Moon}&=mg_{Moon} \\\\ &=2.5\times 1.6 \\\\ &=\boxed{4\,\rm N}\end{align*} Note that the SI units of mass and weight are $\rm kg$ and $\rm N$, respectively. (d) The only consequence of applying forces to an object is a change in its velocity. On the other hand, the straight distance between the force action point and the pivot point is $r=L$. Two forces; upward tension, and downward weight are acting on the body. A $1-\rm {kg}$ bird sits on the midpoint of the rope so that sag of $12^\circ$ is formed. (a) A force $F$ is applied to the left end perpendicular to the radial line $r$, such forces create maximum torque whose magnitude is \[\tau_a=rF=\boxed{4L}\] (b) In this case, the force $F$ is applied perpendicularly to the middle of the radial line, so the distance between the force action point and the pivot point is $r=\frac L2$ \[\tau_b=rF=4(\frac L2 )=\boxed{2L}\] (c) Here, the line of action of the force makes a $45^\circ$ angle with the radial line, $\theta=45^\circ$. The student should be able to (a) state and explain Newton's law of inertia (1st law of motion) and, (b) describe inertia and its relationship to mass. At this point, these two forces, equal in magnitude but opposite in direction, form as shown in the figure below. Note: Due to recent changed in the AP Curriculum from College Board, the order of testing can vary in this class. f m m v v 0 m = mass 1 2 1 1 2 2 m m m x m x xcm. The other torques are \begin{align*} \tau_1&=rF\sin\theta \\&=(1)(55) \sin 66^\circ \\&=50.24\quad \rm m.N \\\\ \tau_2&=rF\sin\theta \\&=(1)(40) \sin 27^\circ \\ &=18.16\quad \rm m.N\end{align*} The forces $F_2$ and $F_1$ rotate the rod about point $C$ in a counterclockwise direction, so by sign conventions for torques, a positive sign must be assigned to them. Applying Newton's second law, we have \[ W_{2x}-W_{1x}-f_{k1}-f_{k2}=(m_1+m_2)a\] where $f_k$'s are the kinetic frictions and are defined as $f_k=\mu_k F_N$. The magnitude of each torque is calculated by the general torque equation as below \begin{align*} \tau_1&=rF\sin\theta \\&=\mathcal l_1 (mg) \sin 90^\circ \\&=\mathcal l_1 mg \\\\ tau_2&=rF\sin\theta \\&=\mathcal l_2 (mg) \sin 90^\circ \\&=\mathcal l_2 mg \end{align*} The net torque about the pivot point is the sum of the torques due to the applied forces: \begin{align*} \tau_{net}&=\tau_1+\tau_2 \\&=+\mathcal l_1 mg + (-\mathcal l_2 mg) \\ &=mg( \mathcal l_1-\mathcal l_2) \end{align*} In the last step, $mg$ is factored out. According to Newton's third law, the force that both masses exerted on each other is the same in magnitude but opposite in direction. 63437 Comments Please sign inor registerto post comments. Due to Newton's first law of motion, when the force is applied abruptly to the lower thread, the hanging block at the other end is still at rest and wants to remain in this situation. answer choices an object wants to maintain its motion if the forces are balanced, then the velocity will change a block will accelerate if a force acts upon it. Find the net vertical force pushing up on the object at this point of the circular path. (c) 2.5 , 1.44 (d) 2.5 , 4. We and our partners use cookies to Store and/or access information on a device. AP Physics 1 Dynamics Free Response Problems ANS KEY 1. In the following, we are going to practice some simple problems about torque to deepen our understanding of these concepts. Practice Problem (16): In the following figure, What are the normal forces at the surfaces of $A$, $B$, and $C$ in $\rm N$, respectively? Physics problems and solutions aimed for high school and college students are provided. Learning Opportunities for AP Coordinators, AP Physics 1: Algebra-Based Past Exam Questions. ins.style.display = 'block'; The force $F_A$ rotates the rod with respect to point $O$ counterclockwise, so its corresponding torque is positive with a magnitude of \begin{align*} \tau_A&=r_AF_A\sin\theta \\&=5\times 12\times \sin 90^\circ \\ &=60\quad \rm m.N \end{align*} On the other hand, the force $F_B$ tend to rotate the rod about $O$ clockwise, so we assign a negative to its corresponding torque magnitude, \begin{align*} \tau_B&=r_BF_B\sin\theta \\&=3\times 8\times \sin 37^\circ \\ &=14.4\quad \rm m.N \end{align*} When more than one torque acts on an object, the torques are added and gives the net torque exerted on the object. We and our partners use data for Personalised ads and content, ad and content measurement, audience insights and product development. One longer way is, first, to find the car's acceleration then use the equation v=v_0+at v = v0 +at and solve for t t. Another much shorter way, which suitable for AP Physics kinematics practice Problems, is using the formula below \Delta x=\frac {v_1+v_2} {2}\Delta t x = 2v1 +v2t . \begin{gather*} F_{Px}=F_P \cos 37^\circ \\\\ F_{Py}=F_P\sin 37^\circ \end{gather*} Apply Newton's second law to the forces along the vertical direction and solve for $F_N$ as below \begin{align*} \Sigma F_y&=ma_y\\\\ F_N+ F_{Py}-mg&=0 \\\\ \Rightarrow F_N&=mg-F_P \sin 37^\circ \\\\ &=(2\times 10)-25 (0.6) \\\\ &=\boxed{5\,{\rm N}}\end{align*}. Each is pulling with a horizontal force. if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-narrow-sky-1','ezslot_14',136,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-narrow-sky-1-0'); Next, find the angle $\theta$ between the force $\vec{F}$ and the line connecting the point of application of the force and the pivot point, which is called the radial line, or position vector $\vec{r}$ in your textbooks. Three force vectors are given and asked for acceleration. The multiple-choice section consists of two question types. (b) Once the applied force is resolved into its radial $F_{\parallel}$ and perpendicular $F_{\bot}$ components, the $F_{\bot}$ points in the counterclockwise direction, so it exerts a positive torque by our sign convention. The torque $tau_1$ acts to rotate the rod clockwise, so a negative is assigned to it. AP Physics 1- Torque, Rotational Inertia, and Angular Momentum Practice Problems FACT: The center of mass of a system of objects obeys Newton's second law: F = Ma cm. Test your knowledge of the skills in this course. Assume $\mu_s=0.4$ and $g=10\,{\rm m/s^2}$. Solution: two equal masses are standing on a level rod pivoted at a point. Problem (12): A $400-{\rm g}$ object releases from a nearly high height. Solve more kinematics questions to master this topic.if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[250,250],'physexams_com-leader-2','ezslot_8',134,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-leader-2-0'); Problem (9): In the figure below, an object is hung from a massless thread. Here, the distance between the point at which the force acts and the nut (axis of rotation) is $r=0.25\,\rm m$. Sign in|Report Abuse|Print Page|Powered By Google Sites, ap-physics-data-analysis-student-guide.pdf, Current Through and Voltage Across Circuit Problems.pdf, series_parallel_circuits_worksheet_02.doc, 1. This torque, due to a frictional force, opposes the overall rotation of the wheel, which is counterclockwise, so it must be supplied by a positive sign, i.e., $\tau_f=+0.3\,\rm m.N$. (take $g=9.8\,{\rm m/s^2}$), (a) 9820 (b) 1250 A total of 769 challenging questions that are divided by topic. As you can see from this statement, the object has to be at rest or moving at a constant speed in order to apply the first law. To view the purposes they believe they have legitimate interest for, or to object to this data processing use the vendor list link below. Single-select questions are each followed by four possible responses, only one of which is correct. In the horizontal direction, there are only two identical components of tension, but in opposite directions. Solution: The weight of an object is defined as $W=mg$ where $g$ is the acceleration of gravity on the surface of a planet. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators . $mg\sin\theta$ down the incline, the normal force $N$, $mg\cos\theta$, and external force $F$ perpendicular to the incline, and finally the static friction force which is the direction must be determined. Calculate the force. (c) $-7$ (d) $-1.3$. Balancing the forces along the vertical and horizontal directions gives us \begin{gather} T_1 \sin 37^\circ=mg \\ T_1 \cos 37^\circ=T_2 \end{gather} Dividing the first expression by the second, the tension $T_1$ cancels out, and we have left the tension $T_2$ as below \begin{align*} T_2&=\frac{mg}{\tan 37^\circ} \\\\ &=\frac{600}{0.6/0.8}\\\\&=\boxed{800\quad {\rm N}}\end{align*} where we used the relation below \[\tan 37^\circ=\frac{\sin 37^\circ}{\cos 37^\circ}\] Substitute $T_2=800\,{\rm N}$ into the second equation $(2)$ and solve for $T_1$ as below \begin{align*} T_1&=\frac{T_2}{\cos 37^\circ}\\\\ &=\frac{800}{0.8}\\\\&=\boxed{1000\quad {\rm N}} \end{align*} Hence, the correct answer is (a). Problem (2): Two forces ($F_A=12\,\rm N$ and $F_B=8\,\rm N$) are applied to a $5-\rm m$ stick as in the figure below. The weight on Mars is given, so we can find the mass of the object \[m=\frac{W_{Mars}}{g_{Mars}}=\frac{9}{3.6}=2.5\,{\rm kg}\] Notice that the mass of any object is constant everywhere, regardless of where it is located. Therefore, we have \begin{align*} 2T\cos\theta&=mg \\\\ \Rightarrow T&=\frac{mg}{2\cos\theta}\\\\&=\frac{60\times 10}{2\cos 37^\circ}\\\\&=\boxed{375\quad{\rm N}}\end{align*} Hence, the correct answer is (c). First, calculate the magnitude of torques associated with each mass exerted on the rod, then assign a positive or negative sign to each torque to indicate their direction. 2020 Exam SAMPLE Question 1 (Adapted from: AP Physics 1 Course and Exam Description FRQ 1) Allotted time: 25 minutes (+ 5 minutes to submit) A small sphere of mass . Problem (11): A mechanic is loosening a nut using a $25-\rm cm$-long wrench by applying a force of $20\,\rm N$ at an angle of $30^\circ$ to the end of the handle. Hence, the total torque with respect to the point $O$ is \[\tau_t=-1+(+0.3)=\boxed{-0.7\,\rm m.N}\]if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-large-mobile-banner-2','ezslot_7',133,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-large-mobile-banner-2-0'); Problem (5): A person exerts a force of $50\,\rm N$ on the end of an $86-\rm cm$-wide door to open it. Thus, the correct answer is c . To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Our mission is to provide a free, world-class education to anyone, anywhere. Problem (9): Calculate the net torque (magnitude and direction) applied to the beam in the following figure about (a) the axis through point $O$ perpendicular to the page and (b) the point $C$ perpendicular to the plane of the page. Thus, the correct choice is (c). The net force of these two gives an upward acceleration to the object. Select a chapter and click on practice questions., AP Physics 1 | Practice Exams | Free Response | Notes | Videos |Study Guides. In the following figure, the forces are resolved into $F_{\parallel}$ and $F_{\bot}$. Vector fields Fundamental forces Gravitational forces Gravitational fields and acceleration due to gravity on different planets Centripetal acceleration and centripetal force Free-body diagrams for objects in uniform circular motion Applications of circular motion and gravitation Energy and momentum 0/500 Mastery points Solution: First, draw a free-body diagram and label all forces acting on the crate as shown below. AP Physics 1: Algebra-Based Past Exam Questions - AP Central | College Board AP Physics 1: Algebra-Based Past Exam Questions Free-Response Questions Download free-response questions from past exams along with scoring guidelines, sample responses from exam takers, and scoring distributions. The frame of reference of any problem is assumed to be inertial unless otherwise stated. Hence, the correct answer is (d). Considering the rod is held initially in the horizontal position and released, what is the net torque (magnitude and direction) on the pivot when it is just released? A 5 kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N. opposing the motion. All forces questions on the AP Physics 1 exams, cover one of the following subsections: Newton's First law Problem (1): In the figure below, we first gently pull the thread down and gradually increase this force until one of the threads connected to the hanging block becomes torn. Balancing the forces exerted on $m_2$ first, gives us \begin{align*} N_{12}-m_2g&=0 \\ N_{12}&=m_2g\\ &=5\times 10 \\&=\boxed{50\,{\rm N}}\end{align*} Thus, the normal force exerted on $m_2$ by the bottom box of $m_1$ is $50\,{\rm N}$. Here, we set the final velocity zero, $v=0$, since we want the maximum distance the block moves up. Assuming the student has worked hard, a student should expect to make a sufficiently high score on the College Board . Solution: Newton's second law of motion has two mathematical forms; one is $\vec{F}_{net}=m\vec{a}$, and the other is $\vec{F}_{av}=\frac{\Delta \vec{P}}{\Delta t}$. This book is Learning List-approved for AP(R) Physics courses. The multiple-choice section consists of two question types. [See Science Practice 1.4] Learning Objective (4.C.2.1): The student is able to make predictions about the . Forces with 3 objects. If you are using assistive technology and need help accessing these PDFs in another format, contact Services for Students with Disabilities at 212-713-8333 or by email at [emailprotected]. Problem (11): Which of the following velocity vs. time graphs below has a correct description for the rain droplet of the previous problem? We again repeat this experiment, but this time, the thread is pulled abruptly down so that one of the threads breaks. Physexams.com, AP Physics 1 Forces Practice Problems + Sample MCQs, 11 Interesting Facts about Gravity | Examsegg. Inertia and Newton's 1st law of motion. The elevator starts moving down initially at rest. where . Combining all these and substituting the numerical values, the frictions and parallel incline weight components are determined as \begin{align*} f_{k1}&=\mu_k m_1g\sin\theta_1\\ &=(0.3)(2)(10) \sin 53^\circ\\&=4.8\,{\rm N} \\\\ f_{k2}&=\mu_k m_2g\sin\theta_2\\ &=(0.3)(5)(10) \sin 37^\circ\\&=9\,{\rm N} \\\\ W_{1x}&=m_1g\sin\theta_1\\ &=(2)(10) \sin 53^\circ \\&=16\,{\rm N} \\\\ W_{2x}&=m_2g\sin\theta_2\\ &=(5)(10) \sin 37^\circ \\&=30\,{\rm N} \end{align*} Now, put these values into Newton's 2nd law written above, \begin{gather*} W_{2x}-W_{1x}-f_{k1}-f_{k2}=(m_1+m_2)a \\\\ 30-16-4.8-9=(2+5)a \\\\ \Rightarrow \quad a=0.028 \quad {\rm m/s^2}\end{gather*} Thus, the acceleration is closest to (a). One is using the lever arm concept and applying the torque formula, $\tau=r_{\bot}F$, and the other is using the force components, in which only the perpendicular component creates a torque about an axis, $\tau=rF_{\bot}$. Hence, the magnitude of the torque about the axis of rotation $O$ is found as \begin{align*} \tau&=(L\sin\theta)F \\ &=(4\sin 60^\circ)(10) \\&=20\sqrt{3}\quad\rm m.N \end{align*}. Thus, the air resistance also increases uniformly. Positive work is done by a force parallel to an object's displacement. (a) 14000 N (b) 50400 N (a) $x=2\sqrt{t}$ (b) $x=-10t^2+2t$ Common Core Standards Science Literacy. Balancing the forces along the $x$ axis gives us the normal force exerted on the box by the wall \[N=F\] The box is to be at rest, so the box's weight must be balanced with the maximum static friction force. Look for the newest edition of this title, The Princeton Review AP Physics 1 Prep, 2023 To log in and use all the features of Khan Academy, please enable JavaScript in your browser. You can choose to review with the whole set or just a specific area. The AP Physics 1 Exam consists of two sections: a multiple-choice section and a free-response section. In a free-body diagram, draw and label each force. A great way to review topics and then test your comprehension. Substituting the numerical values into it, we obtain the minimum force value for which the block is on the verge of motion. Problem (6): Three forces of $\vec{F}_1=20\hat{i}-50\hat{j}$, $\vec{F}_2=10\hat{i}+20\hat{j}$, and $\vec{F}_3=-10\hat{i}$ are acting on a $5-{\rm kg}$ object simultaneously. Summing the corresponding components gives the components of the net force as below \[\vec{F}_{net}=30\hat{i}-40\hat{j}\] The magnitude of this force vector is found as \[F_{net}=\sqrt{30^2+(-40)^2}=50\,{\rm N}\] Dividing the net force by the object's mass gives the acceleration \[a=\frac{F_{net}}{m}=\frac{50}{5}=10\,{\rm m/s^2}\] Hence, the correct answer is (c). AP Physics 1: Electrical Forces and. Comments. An object is moving at 50 . After striking the ground it rebounds at a height of $15\,{\rm m}$. Problem (30): A $3-{\rm kg}$ box has been held fixed on a $30^\circ$ incline by an external force,$F$, perpendicular to it. (a) How far up the incline will it go? Thus, the reaction force is down or $\vec{W}$. \[|a_U|>|a_D|\] Hence, the correct answer is (b). What is the mass of the object and its weight on the surface of the Moon in SI units? (c) 24 N (d) 50 N. Solution: To the box, the following forces are applied. Princeton Review AP Physics 1 Prep, 2022 - The Princeton Review 2021-08-03 Make sure you're studying with the most up-to-date prep materials! . Problem (3): An automobile moves along a straight road at a constant speed. (b) in this part, the angle between $r$ and $F$ is $\theta=53^\circ$ as illustrated in the figure below. Unit 11 Practice Problems. container.style.maxHeight = container.style.minHeight + 'px'; Because it is possible some forces are applied to an object at rest and the object stays at rest or in another situation, those forces are applied to a constant speed moving object but the object's velocity does not change. \[F=\frac{2\times 10}{0.4}=50\,{\rm N}\], Problem (19): A block of mass $m=10\,{\rm kg}$ is hung from two identical strings which makes an angle of $37^\circ$ with the vertical. (c) $x=10t$ (d) $v=-10t+3$. The individuals who are preparing for Physics GRE Subject, AP, SAT, ACTexams in physics can make the most of this collection. These online tests include hundreds of free practice questions along with detailed explanations. (Take $g=10\,{\rm m/s^2}$). Now, using the formula $F_{net}=ma$, we can find the average force that is required to stop this car as below \[F=3500\times 4=\boxed{14000\,{\rm N}}\] Hence, the correct answer is (a). (a) $\vec{W}$,$\vec{W}$ (b) $-\vec{W}$,$\vec{W}$ On the diagram of the block below, draw and label all the forces that act on . The downward force is also the force exerted by the thread on the ceiling and pulls it down. Newton's third law and free body-diagrams, Gravitational fields and acceleration due to gravity on different planets, Centripetal acceleration and centripetal force, Free-body diagrams for objects in uniform circular motion, Applications of circular motion and gravitation. AP Physics 1- Dynamics Practice Problems ANSWERS FACT: Inertia is the tendency of an object to resist a change in state of motion. What is the magnitude of the torque if the force is applied (a) perpendicular to the door and (b) at an angle of $53^\circ$ to the plane of the door? Solution: According to Newton's second law, a net force applied to an object can accelerate it by $a=\frac{F_{net}}{m}$. What air resistive force is applied to the car? (c) In the first experiment, the upper thread breaks but in the second the lower thread. (a) What torque does the mechanic apply to the center of the nut? This is the force that is responsible for pulling the box down and accelerating it. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Sort by: Top Voted Problem (26): A person weighing $60,{\rm kg}$ stands on a scale in a moving elevator. (b) To find the torque of this configuration, extend the force $F$ and draw a line perpendicular to it so that it passes through the axis of rotation. All content of site and practice tests copyright 2017 Max. You push the box against the wall with a force of $F$ rightward. (c) Again, identify the lever arm and compute the magnitude of the torque associated with this force about point $O$. window.ezoSTPixelAdd(slotId, 'adsensetype', 1); Generate a 10 or 20 question quiz from this unit and find other useful practice. Thus, the torque associated with this force is found to be \begin{align*} \tau&=rF\sin\theta \\&=(0.86)(50) \sin 53^\circ \\ &=34.4\quad \rm m.N\end{align*} From this torque question, we can understand the physical concept of torque. The forces $F_2$ and $F_3$ rotate the rod about the point $Q$ in ccw and cw directions, respectively, resulting in a positive and negative torque. Examples of scalar quantities are mass, time, area, temperature, emf, electric current, etc. var slotId = 'div-gpt-ad-physexams_com-medrectangle-3-0'; Problem (3): Calculate the net torque about the axle of the wheel through point $O$ perpendicular to the plane of the page, taking $r=12\,\rm cm$ and $R=24\,\rm cm$. Created by David SantoPietro. The only force along the incline is the component of the weight downward, $mg\sin\theta$. This course is equivalent to a first-year/first semester calculus-based classical mechanics college physics class and is designed to prepare students for the AP Physics C Mechanics Exam given in May. Force: Force & Mass The following circular motion questions are helpful for the AP physics exam. How many times is the force that $m_1$ exerts on $m_2$ than the force exerted on the surface by $m_1$? Thus, the only force that is exerted on the block is $W_x=mg\sin\theta$ down the incline. var pid = 'ca-pub-8931278327601846'; Therefore, only choice (c) has the form of a motion in which the object moves at a constant speed. (a) 25 (b) 30 Student resources for Physics: Algebra/Trig (3rd Edition) by Eugene Hecht. p = momentum . D. During the collision, the truck has a greater . The forces $F_1$ and $F_2$ rotate the wheel clockwise, which exerts negative torques on the wheel whose magnitudes are found as follows \begin{align*} \tau_1&=r_{\bot,1}F_1 \\&=(0.20)(15) \\&=3\quad \rm m.N \\\\ \tau_2&=r_{\bot,2}F_2 \\&=(0.20)(10) \\&=2\quad \rm m.N \end{align*} The other force $F_3$ that acts at an angle with the rime of the smaller circle apply a positive torque according to the sign conventions for torques (counterclockwise rotation). The force $F_1$ rotates the smaller circle with the lever arm $r_{\bot,1}=0.12\,\rm m$ clockwise, so assign a negative to its torque magnitude. The coefficient of kinetic friction is k, between block and surface. 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