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Acceleration of Masses - Introductory Physics - Solved Paper, Exams of Physics

These are the notes of Solved Paper of Introductory Physics and its key important points are: Acceleration of Masses, Intersecting Roads, Unit-Vector Notation, Kinetic Friction, Acceleration of Masses, Scale Read, Curvature of Track

Typology: Exams

2012/2013

Uploaded on 02/12/2013

sathianarayan
sathianarayan 🇮🇳

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Download Acceleration of Masses - Introductory Physics - Solved Paper and more Exams Physics in PDF only on Docsity! Physics 8A Summer 2002 MIDTERM 1 Each problem is worth 10 points. 1. A ball is thrown from the ground into the air, reaching a maximum height of 115 m, and traveling a horizontal distance of 245 m before it hits the ground. Determine the speed and direction with which the ball was thrown, and-the amount of time it was in the air. + 2. Acar a bus travel along two intersecting roads, as Ns shown here. is traveling at 30 m/s due north, and Covered the bus is traveling at 20 Tn/s-at 35° north of east. At t= 0, the bus is at the intersection, while ar is 75 m tothe Fe | { south of the intersection. (a) Find the veloct en L co 3 (You should take no: firection of the +x axis.) (b) Find the ¢ Car relative to the bus as a function of ‘expressed in unit-vector notation. tim 3. In the diagram, m, = 2.0 kg and m, = 3.0kg. A coefficient of kinetic friction w= 0.30 exists between both blocks and the surfaces. The pulley is massless and frictionless. Find the acceleration of the masses, and the tension in the rope. 4. (a) A roller coaster car travels at 25.0 m/s at the bottom of a loop, where the radius of curvature is 65.0 m. An 85.0 kg man is in the car, and sitting on a scale. What does the scale read? (b) Suppose the car slows to 15.0 m/s at the top, and the scale now reads an amount equal to the man’s weight. What must be the radius of curvature of the track at the top? Physics FA Mi dkhern 1 Salaton, | Bt fk ee tee, The on weg oun Con be frunk Frere! YrYet Mey * mk she Lu “No = o-bauty t+, 7 UPy, This Oy ele thn BR yo ue? te, z4y 2 4. Fay ro HEL tem Fe dt= Fors Now 6 Vox Xm Let Van t LUO 2 Wy CU CFI Von 7 LSP Te furs Vey Vy 7 Vey ~5 hey O * Vey —@ WEY, FY) Vey > 414 6 Ve Vox t* Vay > 532.7 \aal Aa tan G. = Voy 2 17 Vox. 6,~- 619°
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