The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, what will the pressure of a gas be after the temperature changes?
Answer: Well the answer is KE = 5.625E-7 i just don't know the units for it...
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To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.
For this purpose we have that the constructive interference in waves can be expressed under the function

Where
a = Width of the slit
d = Distance of slit to screen
m = Number of order which represent the number of repetition of the spectrum
Angle between incident rays and scatter planes
At the same time the distance on the screen from the central point, would be

Where y = Represents the distance on the screen from the central point
PART A ) From the previous equation if we arrange to find the angle we have that



PART B) Equation both equations we have


Re-arrange to find a,


Answer:
The ball will have a kinetic energy of 0.615 Joules.
Explanation:
Use the kinetic energy formula

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)