So there is a question at the end of the fifth passage in PS in the TPR Diagnostic #2 that says:
If air is moving through the cylinder and y is 4cm what is the speed of air?
The answer is 23m/s
The explanation for the answer says:
applying the equation given in the passage yields:
v= the sq. root of (2(10m/s^2)(4x10^-2m)(8x10^2kg)/ (1.2kg/m^3))
So i can't find any equations in this passage and i'm just completely confused, where is the 8x10^2 coming from? or the 1.2kg/m^3? please help!
If air is moving through the cylinder and y is 4cm what is the speed of air?
The answer is 23m/s
The explanation for the answer says:
applying the equation given in the passage yields:
v= the sq. root of (2(10m/s^2)(4x10^-2m)(8x10^2kg)/ (1.2kg/m^3))
So i can't find any equations in this passage and i'm just completely confused, where is the 8x10^2 coming from? or the 1.2kg/m^3? please help!
