1. golf ball is driven horizontally from an elevated tee with velocity of 80 ft sec it strikes the fairway 2.5sec later . how far has it travelled horizontally?

2. A 0.064 kg of octane vpor ( MW=114) is mixed with 0.91 kg of air (MW=29.0) in manifold of an engine. The total pressure in the manifold is 86.1 KPA and a temprature is 290 K. Assume octane behaves ideally . What is the partial pressure of the air in mixture in kPa.

Question

1. golf ball is driven horizontally from an elevated tee with velocity of 80 ft  sec it strikes the fairway 2.5sec later . how far has it travelled horizontally?

2. A 0.064 kg of octane vpor ( MW=114) is mixed with 0.91 kg of air (MW=29.0) in manifold of an engine. The total pressure in the manifold is 86.1 KPA and a temprature is 290 K. Assume octane behaves ideally . What is the partial pressure of the air in mixture in kPa.

osprey · Answer

Here's my guess... speed=distance/time -> distance=speedxtime -> 80x2.5=200feet ~ 70 yards. Amateur golfer ? http://perendis.webs.com/remembranceday/index.htm ********* http://www.perendis.webs.com/RELATIVITY/RELATIVITY%20index.htm ***** http://www.perendis.webs.com *********

osprey · Answer

Here's my first guess...
Application of Dalton's law of partial pressures. 
Octane partial pressure = mole fraction of octane times the total pressure in manifold.
Mole fraction is number of moles of octane divided by total number of moles present.
Could need to convert the mass of the gases given into moles.

Done in both physics and chemistry. Chem seems to go into more detail.