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Common Acceleration of Objects - Physics for Scientist and Engineers - Solved Past Paper, Exams of Engineering Physics

This is the Solved Past Paper of Physics for Scientist and Engineers which includes Coaxial Cable, Cylindrical Tube, Water Temperature, Single Loop etc. Key important points are: Common Acceleration of Objects, Corresponding Coefficient, Coefficient of Friction, Inverted Cone, Terminal Velocity, Positive Constant, Parabolic Path

Typology: Exams

2012/2013

Uploaded on 02/12/2013

sashekala
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Download Common Acceleration of Objects - Physics for Scientist and Engineers - Solved Past Paper and more Exams Engineering Physics in PDF only on Docsity! 1. (20 points) Two objects of masses m, = 1.65 kg and mg = 3.22 kg attached by a massless rod parallel to the incline on which both slide travel down the plane with m; trailing ma, The angle of the incline is 6 = 29,5°. The coefficient of friction between m, and the incline is 41 = 0.226; between m; and the incline the corresponding coefficient is ba = 0.127. QO {a) Compute the common acceleration of the two objects. (b) Determine the tension in the rod. (c} What are the answers to (a) and (b) if my trails m,? (a) m, mg om & tT — MGI cos & nya in, mn, gen -T ~My mg essk = wa > 7, 5 f . 1 tri) > (m, + m, ) game LAM “hae | Ge sb Cat, 4 ui ' i, >| 2 a2 US ne hes | B.456 m/s | ‘ ee i tan bj = ; > { &. if eo () Ts mosh ~ my teat PL a = My, (gsm - 4, g oe. “) 0-42. My («¢) ™, : mm, gain -T _ AY my q coy - WY vy, a, gen a Ay gent = & = ty na 7 . > (mH 4m,) gin ~ (gm + pe) genG 2 ON Te => | Same a and T] T “ee he. t Ye direction 2. (20 points) A small block of mass m is placed inside an inverted cone that is rotating about a vertical axis such that the time period for one revolution is P. The walls of the cone make an angle 8 with the vertical. The coefficient of static friction between the block and the cone is ps. If the block is to remain at a constant height h above the point of the cone, what are the maximum and minimum values of P? y R= h tan 6 cD y r jort before it slips, Pop, A 2 x X- Component: Foe aes Neos 9- Fsin@ total Agtance . 2T R Ve latel tyme P m( suf)” = Noos @ ~ Ks 806 a ale R ae . HT? (4 tan 8) = wv (cas d- A 56) p* iad yn Component : Mend 4 Fesd -m4g = 0 fv (8108 + As cos @) = m4 =? N= ed Sind ts sd > Antmb 6 = m9 ) (cers -vs8) “pe tan 3:20 ts ord meer ) Panay > AT fiboa? | T= Ps tan? 4 = 9° : Yath S00 tA sO => Rene tar) tLe tang as 8-4, 5:08 Pain = aT 9 epiend|
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