First overtone of open organ pipe is given as

first overtone of closed organ pipe is given as

now they are in unison so we will have




so end correction of both pipes is e = 1 cm
The statement can't be true. Objects with different masses held at the same height don't have the same gravitational potential energy.
The density of seawater at a depth where the pressure is 500 atm is 
Explanation:
The relationship between bulk modulus and pressure is the following:

where
B is the bulk modulus
is the density at surface
is the variation of pressure
is the variation of density
In this problem, we have:
is the bulk modulus

is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)
Therefore, we can find the density of the water where the pressure is 500 atm as follows:

Learn more about pressure in a fluid:
brainly.com/question/9805263
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D. <span>Johannes Kepler argued that Earth was the center of the universe.
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His is a step down transformer since n(primary) is greater than n(seconcary). You relate the input voltage with the ouput voltage with the following equation:
<span>Vout = n2/n1*Vin (n2/n1 is essentially your 'transfer function' that dictates what a specified input would produce) </span>
<span>Solving the equation: </span>
<span>Vin = Vout*n1/n2 = (320V)*(600/300) = 640 V </span>
<span>This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.</span>