A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.50 s, it is at
point (5.80 m, 5.60 m) with velocity (3.50 m/s) j and acceleration in the positive x direction. At time t2 = 11.7 s, it has velocity (–3.50 m/s) i and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.Expert Answer
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Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
which can also be rewritten as
(1)
where the labels 1 and 2 mark the initial and final conditions of the gas.
In our problem,
, 
and 
, so the final pressure of the gas can be found by re-arranging eq.(1):
Therefore the correct answer is
<span>1. 0.75 atm</span>
which can also be rewritten as
where the labels 1 and 2 mark the initial and final conditions of the gas.
In our problem,
Therefore the correct answer is
<span>1. 0.75 atm</span>