Here it is. *WARNING* VERY LONG ANSWER
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11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h)
The change in PE =mgh=5*9.8*12=588 J
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12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s.
Commendable and jockey Pat Day had a combined mass =M= 550.0 kg,
Their KE as they crossed the line=(1/2)Mv^2
Their KE as they crossed the line=0.5*550*(15.98)^2
Their KE as they crossed the line is 70224.11 J
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13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m
She trips and drops the spare tire of mass = m = 10.0 kg,
The tire rolls down the hill with an intial speed = u = 2.00 m/s.
The height of top of the next hill = h = 5.00 m
Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2
Initial total mechanical energy =mgH+(1/2)mu^2
Suppose the final speed at the top of second hill is v
Final total mechanical energy =PE+KE=mgh+(1/2)mv^2
As mechanical energy is conserved,
Final total mechanical energy =Initial total mechanical energy
mgh+(1/2)mv^2=mgH+(1/2)mu^2
v = sq rt [u^2+2g(H-h)]
v = sq rt [4+2*9.8(20-5)]
v = sq rt 298
v =17.2627 m/s
The speed of the tire at the top of the next hill is 17.2627 m/s
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14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean.
a.)The mass of bean = m = 2.0 g
Height up to which the been jumps = h = 1.0 cm from hand
Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg
b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s
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15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill.
The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s
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16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m,
The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2
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EDIT
1.) A train is accelerating at a rate = a = 2.0 km/hr/s.
Acceleration
Initial velocity = u = 20 km/hr,
Velocity after 30 seconds = v = u + at
Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s =
Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr
Velocity after 30 seconds = v = 80 km/hr=22.22 m/s
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2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins.
His acceleration = a =11.1/9=1.233 m/s^2
Distance he covered = s = (1/2)at^2=49.95 m