28) s = ut + 0.5*at2 S = 12 m a = 9.8 m/s2 u = 0 implies t = 1.5 sec . For cadet velocity = 6 km/hr = 1.6 m/sec S = vt = 1.57 × 1.66 = 2.6 m.
22) v1 = 16.6 m/s.v2 = 11.6 m/s.Relative velocity between the cars = (16.6 – 11.6) = 5 m/s.Distance to be travelled by first car is 5 + 5 = 10 m.Time = t = s/v = 10/5 = 2 sec to cross the 2nd car.In 2 sec the 1st car moved = 16.6 × 2 = 33.2 mH also covered its own length 5 m.Therefore total road distance used for the overtake = 33.2 + 5 = 38 m.
21) VP = 25 m/s.VC = 20 m/s.In 10 sec culprit reaches at point B from A.Distance converted by culprit S = vt = 20 × 10 = 200 m.At time t = 10 sec the police jeep is 200 m behind theculprit.Time = s/v = 200 / 5 = 40 s. (Relative velocity).In 40 s the police jeep will move from A to a distance S, whereS = vt = 25 × 40 = 1000 m = 1.0 km away.
21) From the free body diagramR + T – Mg = 0implies R = Mg – T ...(i)R1 – R – mg = 0implies R1 = R + mg ...(ii)And T – (nu) R1 = 0 , T – (nu) (R + mg) = 0 [From equn. (ii)] T – (nu) R – (nu) mg = 0 T – (nu) (Mg + T) – (nu) mg = 0 [from (i)] T (1 + (nu)) = (nu)Mg + (nu) mg thereforeT = (nu)(M+m)g/1+(nu)
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35) m1 = 0.1kgm2 = 0.5kgm3 = 0.05kg.T + 0.5a – 0.5g = 0 ...(i)T1 – 0.5a – 0.05g = 0 ...(ii)T1 + 0.1a – T + 0.05g = 0 ...(iii)From equn (ii) T1 = 0.05g + 0.05a ...(iv)From equn (i) T1 = 0.5g – 0.5a ...(v)Equn (iii) becomes T1 + 0.1a – T + 0.05g = 0implies 0.05g + 0.05a + 0.1a – 0.5g + 0.5a + 0.05g = 0 [From (iv)and (v)] implies 0.65a = 0.4g therefore a = 8g/13 downward
go to http://chemistry.boisestate.edu/people/richardbanks/inorganic/bonding%20and%20hybridization/bonding_hybridization.htm & http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swf
31) Ma – 2T = 0 implies T = Ma /2. , T + Ma – Mg = 0 implies Ma/2 + Ma = Mg. (because T = Ma/2) implies a = 2g/3 a) acceleration of mass M is 2g/3. ,b) T = Ma /2 therefore a = Mg/3 , c) Let, R1 = resultant of tensions = force exerted by the clamp on the pulley , therefore R1 = root(T2 + T2) = (root2)T therefore R = (root2)T implies (root2)Mg/3 As, Tan$ =T/T =1 therefore $ =45(degree) , So, it is (root2)Mg/3 at an angle of 45(degree) with horizontal
20) When the box is accelerating upward,U – mg – m(g/6) = 0 (U = buoyant force)implies U = mg + mg/6 = m{g + (g/6)} = 7 mg/6 …(i)implies m = 6U/7g.When it is accelerating downward, let the required mass be M.U – Mg + Mg/6 = 0 implies U = 6Mg - Mg/6 = 5Mg/6 therefore M = 6U/5g Mass to be added = M – m = 6U/Mg - 6U/7g implies 12/35(7mg/6g) from (i) The mass to be added is 2m/5.
37. Suppose the monkey accelerates upward with acceleration ’a’ & the block, accelerate downward withacceleration a1. Let Force exerted by monkey is equal to ‘T’From the free body diagram of monkeytherefore T – mg – ma = 0 ...(i)implies T = mg + ma.Again, from the FBD of the block,T = ma1 – mg = 0.implies mg + ma + ma1 – mg = 0 [From (i)] implies ma = –ma1 implies a = -a1.Acceleration ‘–a’ downward i.e. ‘a’ upward.Therefore the block & the monkey move in the same direction with equal acceleration.If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as theyhave same weight (same mass). Their separation will not change as time passes
22) q1 = q2 = 2 × 10–7 c , m = 100 gl = 50 cm = 5 × 10–2 m , d = 5 × 10–2 m(a) Now Electric forceF = Kq2/r2 implies 9*109*4*10 -14 / 25*10 -4 N implies 14.4 N = 0.144N(b) The components of Resultant force along it is zero,because mg balances T cos$ and so also.F = mg = T sin$(c) Tension on the stringT sin$ = F , T cos$ = mg implies T2 = F2+(mg)2 implies (0.144)2 + (0.1*9.8)2 implies T = 0.986 NTan$ = F/mg = 0.144 / 100*10 -3*9.8 = = 0.14693But T cos$= 102 × 10–3 × 10 = 1 Nimplies T = 1/cos$ = sec$implies T = F/sin$Sin$ = 0.145369 ; Cos$ = 0.989378
Ionic RadiiThe shortest cation–anion distance in an ionic compound corresponds to the sum of theionic radii. This distance can be determined experimentally. However, there is no straightforwardwayto obtain values for the radii themselves.Data taken from carefully performedX-ray diffraction experiments allow the calculation of the electron density in the crystal;the point having the minimum electron density along the connection line between a cationand an adjacent anion can be taken as the contact point of the ions. As shown in the exampleof sodium fluoride , the ions in the crystal show certain deviations fromspherical shape, i.e. the electron shell is polarized. This indicates the presence of somedegree of covalent bonding, which can be interpreted as a partial backflow of electrondensity from the anion to the cation. The electron density minimum therefore does notnecessarily represent the ideal place for the limit between cation and anion. Ionic radii can also be used when considerable covalent bonding is involved. The higherthe charge of a cation, the greater is its polarizing effect on a neighboring anion, i.e. thecovalent character of the bond increases.
Covalent RadiiCovalent radii are derived from the observed distances between covalently bonded atomsof the same element. For example, the C–C bond length in diamond and in alkanes is 154pm; half of this value, 77 pm, is the covalent radius for a single bond at a carbon atomhaving coordination number 4 (sp3 C atom). In the same way we calculate the covalentradii for chlorine (100 pm) from the Cl–Cl distance in a Cl2 molecule, for oxygen (73pm) from the O–O distance in H2O2 and for silicon (118 pm) from the bond length inelemental silicon. If we add the covalent radii for C and Cl, we obtain 77+100=177 pm;this value corresponds rather well to the distances observed in C–Cl compounds.However,if we add the covalent radii for Si and O, 118+73 =191 pm, the value obtained does notagree satisfactorily with the distances observed in SiO2 (158 to 162 pm). Generally wemust state: the more polar a bond is, the more its length deviates to lower values comparedwith the sum of the covalent radii.
Van der Waals RadiiIn a crystalline compound consisting of molecules, the molecules usually are packed asclose as possible, but with atoms of neighboring molecules not coming closer than thesums of their VAN DER WAALS radii. The shortest commonly observed distance betweenatoms of the same element in adjacent molecules is taken to calculate the VANDER WAALS radius for this element. A more detailed study reveals that covalently bonded atoms are not exactly spherical. For instance, a halogenatom bonded to a carbon atom is flattened to some degree, i.e. its VAN DER WAALSradius is shorter in the direction of the extension of the C–halogen bond than in transversedirections. If the covalent bond is more polar, as in metal halides, then thedeviation from the spherical form is less pronounced. The kind of bonding also can havesome influence; for example, carbon atoms in acetylenes have a slightly bigger radius thanin other compounds.Distances that are shorter than the sums of the corresponding listed values of the VANDER WAALS radii occur when there exist special attractive forces. For example, in a solvatedion the distances between the ion and atoms of the solvent molecules cannot becalculated with the aid of VAN DER WAALS radii. The same applies in the presence ofhydrogen bonding
Inertial frame of referenceAn inertial reference frame is one in which Newton's first and second laws of motion are valid.Hence, within the inertial frame, an object or body accelerates only when a physical force is applied, and (following Newton's first law of motion), in the absence of a net force, a body at rest will remain at rest and a body in motion will continue to move uniformly?ie. in a straight line and at constant speed.Non-inertial reference frame Non inertial frame of reference is one in which a body violates Newton's Laws of Motion, mainly the First Law. In such a frame, despite no real force acting on a body at rest, it might move; or one that was already moving come at rest or change it's direction of motion. For comparison see an inertial frame.Newton's first and second laws of motion do not hold in non-inertial reference frames. Specifically, masses in non-inertial reference frames appear to feel fictitious forces (such as the Coriolis force or the centrifugal force) that derive from the acceleration of the reference frame itself. Fictitious forces cause apparent accelerations in objects without any physical force causing the acceleration. Fictitious forces are proportional to the mass upon which they act; if such forces are observed, scientists will recognize that they are in a non-inertial reference frame. For example, the rotation of the Earth can be observed from the Coriolis force acting on a Foucault pendulum.An apparent exception would seem to be the force of gravity, which is also proportional to the mass upon which it acts. Although gravity can be considered a "real" physical force for the purposes of calculations in classical mechanics, Albert Einstein showed in his theory of general relativity that gravity itself can also be considered a fictitious force. In his theory, the free-falling reference frame is equivalent to an inertial reference frame (the equivalence principle). By contrast, Einstein noted that observers standing on the Earth are experiencing an unrecognized acceleration from the normal force pushing up on their feet and, thus, are in a non-inertial (accelerated) reference frame. Further details may be found under general relativity.
