Find the pressure due to the fluid at a depth of 76 cm in still (a) water (ρw=1.00 g / cm^3) a

step 1 Good day, the topic is about hydrostatic pressure. We note that a certain column of liquid will

Find the pressure due to the fluid at a depth of 76 cm in...

Find the pressure due to the fluid at a depth of $76 \mathrm{~cm}$ in still $(a)$ water $\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)$ and $(b)$ mercury $\left(\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)$. (a) $P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=$ $7.5 \mathrm{kPa}$ (b) $P=\rho g h=(13600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx$ $1.0$ atm

Find the pressure due to the fluid at a depth of $76 \mathrm{~cm}$ in still $(a)$ water $\left(\rho_{w}=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)$ and $(b)$ mercury $\left(\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)$.
(a) $P=\rho_{w} g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=7450 \mathrm{~N} / \mathrm{m}^{2}=$
$7.5 \mathrm{kPa}$
(b) $P=\rho g h=(13600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.76 \mathrm{~m})=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \approx$
$1.0$ atm

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Key Concepts

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth because the weight of the fluid above adds to the pressure felt at a given point, and it is calculated using the formula P = ?gh, where ? is the density of the fluid, g is the gravitational acceleration, and h is the depth.

Fluid Density

Density is a measure of mass per unit volume and is a critical property of a fluid that influences its weight and, consequently, the pressure it exerts at a certain depth. A fluid with higher density will exert more pressure at the same depth compared to a fluid with lower density.

Gravitational Acceleration

Gravitational acceleration, the rate at which objects accelerate under the force of gravity (approximately 9.81 m/s² near the Earth's surface), plays a key role in determining the weight of the fluid column above a point. This acceleration directly affects the hydrostatic pressure, as it multiplies with the fluid's density and depth in the pressure calculation.

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