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This is a straightforward s = (1/2)a t² problem, to ease students into the exam. It's easy to calculate the time to get to Lovelon, and each light year represents

1 light-year = 365 days/year x 24 hours/day x 3600 seconds/hour x 3 x 10⁸ m/s = 9.46 x 10¹⁵ m.

If Romeo travels all the way there at 15 m/s², then 10 light-years will take him a time given by 10 x 9.46 x 10¹⁵ = 1/2 x 15 (time)² giving 1.123 x 10⁸ seconds, or 3.56 years, and he will be moving at 16.85 x 10⁸ m/s, which in Startrek language is Warp 5.6! This won't do him much good of course, because he will pass Lovelon at this speed. Romeo has to travel half-way to Lovelon, 5 light-years, then reverse his rockets to decelerate for another 5 light years at 15 m/s², so that he arrives at low speed for docking. This splits the journey exactly in half, due to the obvious symmetry of the motion. At the half-way point he'll be moving at 11.9 x 10⁸ m/s, which is near enough Warp 4!

It'll take him 7.94 x 10⁷ sec to get half-way, about 2.51 years, so the total time to get to Lovelon and then back to earth will be 4 times this, or 10.07 years. They can wed in 2007 (C).

The acrobat has a force acting on his hand that we resolve into two perpendicular components: the vertical one is the reaction to the weight (115 x 9.8 N = 1127 N) and theÌýhorizontal one balances the 130 N force from the wall. These two forces give a resultant forceÌýF of
F = √(11272 + 1302) = 1134 N (A).

If eachÌýbreath of Sir Isaac's (and ours) is about 1 litre = 10^-3 m³, and they are 3 seconds apart, thenÌýin 1727-1642=85 years, he will have had 85 x 365 x 24 x 3600 /3 = 8.93 x 10⁸ breaths, for aÌýtotal volume of 8.93 x 10⁵ m³. We will assume that the air mixes well enough that we do notÌýhave to worry about air being breathed twice.
The total volume of the atmosphere is 4 R_EÌý2 h, where R_E is the Earth's radius, and h is theÌýheight of the atmosphere, giving 4.08 x 10¹⁸ m³. The fraction of air molecules ever breathedÌýby the patron saint of Physics is thus 8.93 x 10⁵ / 4.08 x 10¹⁸ = 2.19 x 10^-13.
The number ofÌýmolecules in each breath of ours is the density 1/(3.3 x 10^-9)³ = 2.78 x 10²⁵ m^-3, multiplied byÌýthe volume of each breath, 10^-3 m³, or 2.78 x 10²² molecules. Multiplying by the fractionÌýbreathed by Newton, each breath of ours has about 6.08 x 10⁹ molecules also breathed by him (D). Since we have about 9 x 10⁸ breaths, each breath of ours has about 7 molecules also breathed by Isaac Newton.

A free-body diagram shows us that the vertical component of the tension in the two ends ofÌýthe wire, 2Tsin(15°), must equal the total loaded weight W.
The volume of the wire is \(\pi R_{w}2L\), where \(R_{w}\) is the radius of the wire, and the total volume ofÌýice is \(\pi(R_{i}^{2}-R_{w}^{2})L\), where \(R_{i}\) is the radius of the ice-covered wire. Multiplying by theÌýrespective densities, the total mass of the 500m-long ice-covered wire is 694.5 + 3531.5 = 4226 kg, having a total weight of 41415 N. The tension in the wire is thus \(\frac{41425}{2sin15^{\circ}} = 80000 N\), and dividing by the cross-sectional area of 1.77 x 10^-4 m³, the stress is 4.52 x 10⁸Ìý\(\frac{N}{m^{2}}\) (B).

Romeo's flowers are 500 m up, have a downward velocity component of 50 m/s, and they will
strike the surface of the ground in a time given by \(500 = 50t + \frac{9.8t^{2}}{2}\), which solves to
give t = 6.21 seconds. The flowers must cover 40 m horizontally in this interval, so Romeo
must throw them out sideways at \(\frac{40}{6.21} = 6.44 \frac{m}{s}\)Ìý(E).

With each song being 3 minutes long, there are 23.33 songs on a 70-minute CD. This requiresÌýa total storage of 2.9 megabytes x 23.33 = 67.67 megabytes. The modem processes about 30Ìýkilobits/sec, which is 3.75 kilobytes/sec. At this rate, 67.67 megabytes will takeÌý(67.67x10⁶)/(3750) = 18045 seconds or 301 minutes (A) or 5.01 hours!Ìý