Please could you show me how to figure out these problems? Thanks in advance and have a great day.
11.) It is said that Galileo dropped objects off the Leaning Tower of Pisa to determine whether heavy or light objects fall faster. If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, what would have been a change in PE of the cannon ball?
12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed of 15.98 m/s. If Commendable and jockey Pat Day had a combined mass of 550.0 kg, what was their KE as they crossed the line?
13.) Brittany is changing the tire of her car on a steep hill 20.0 m high. She trips and drops the 10.0-kg spare tire, which rolls down the hill with an intial speed of 2.00 m/s. What is the speed of the tire at the top of hte next hill, which is 5.00 m high? (Ignore the effects of rotaion KE and friction.)
14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. a.) If a bean of mass 2.0 g jumps 1.0 cm from your hand into the air, how much potential energy has it gained in reaching its highest point. b.) What is the speed as the bean lands back in the palm of your hand?
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. What is the pig's speed a the bottom of the hill? Use the law of conservation of energy.
16.) While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on Earth. If Neil Armstrong jumped up on the moon with an intitial speed of 1.51 m/s to a height of 0.700 m, what amount of gravitational acceleration did he experience?Please could you help me with these energy physics problems?________________________________________?br>
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
