Which of the following is the full-law calibration equation?

Question

TuteeHUB · Accepted Answer

pp-p=p0$\Bigg[\Big\{1+\frac{\gamma+1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$

TuteeHUB · Answer

pp-p=p0$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$

TuteeHUB · Answer

pp-p=p0$\Bigg[\Big\{1+\frac{\gamma+1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$

TuteeHUB · Answer

p-p0=p0$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$

TuteeHUB · Answer

pp-p=p0$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma+1}}-1\Bigg]$

1.	Which of the following is the full-law calibration equation?
A.	pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)
B.	pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma+1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)
C.	p-p0=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)
D.	pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma+1}}-1\Bigg]\)
Answer» B. pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma+1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)

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