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Vishnu
Vishnu
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Ask yourself why it was pVg$pVg$ in case of non accelerated (inertial) frame. Because the pressure gradient is pg$pg$ or p$p$ times the acceleration due to the force that is pulling it downwards.

Now in an accelerated frame, the pressure gradient is not really pg$pg$ but p(g+a)$p(g+a)$.

If you had to write pressure at a depth d$d$, would you write pVg$pVg$ or pV (g+a)$pV (g+a)$?

Ask yourself why it was pVg in case of non accelerated (inertial) frame. Because the pressure gradient is pg or p times the acceleration due to the force that is pulling it downwards.

Now in an accelerated frame, the pressure gradient is not really pg but p(g+a).

If you had to write pressure at a depth d, would you write pVg or pV (g+a)?

Ask yourself why it was $pVg$ in case of non accelerated (inertial) frame. Because the pressure gradient is $pg$ or $p$ times the acceleration due to the force that is pulling it downwards.

Now in an accelerated frame, the pressure gradient is not really $pg$ but $p(g+a)$.

If you had to write pressure at a depth $d$, would you write $pVg$ or $pV (g+a)$?

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Smit Chaudhary
Smit Chaudhary
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Ask yourself why it was pVg in case of non accelerated (inertial) frame. Because the pressure gradient is pg or p times the acceleration due to the force that is pulling it downwards.

Now in an accelerated frame, the pressure gradient is not really pg but p(g+a).

If you had to write pressure at a depth d, would you write pVg or pV (g+a)?