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3 years ago

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A cube of wood 15.0 cm on each side is tied to the bottom of a tank filled with water to a depth of 50 cm. The tension in the st

ring is found to be 6.615 N. Please answer each of the following questions. Note: 1 atm = 1.013 x 105 Pa a) What is the density of the wood? b) What is the absolute pressure at the bottom of the tank.

1 answer:

Elina [12.6K]3 years ago

7 0

Answer:

(a). The density of the wood is $1479.48\times10^{2}\ Kg/m^3$

(b). The absolute pressure at the bottom of the tank is $1.06200\times10^{5}\ Pa$ .

Explanation:

Given that,

Side of cube = 15.0 cm

Depth = 50 cm

Tension = 6.615 N

We need to calculate the volume of the wood

Using formula of volume

$V = a^3$

$V=(15.0\times10^{-2})^3$

$V=0.003375\ m^3$

We need to calculate the density of the wood

Using buoyant force

$\rho_{w}gh=mg+T$

$\rho_{w}gh=\rho_{c}Vg+T$

Put the value into the formula

$\rho_{c}=\dfrac{\rho_{w}gh-T}{Vg}$

Put the value into the formula

$\rho_{c}=\dfrac{1000\times9.8\times50\times10^{-2}-6.615}{0.003375\times9.8}$

$\rho_{c}=1479.48\times10^{2}\ Kg/m^3$

(b). We need to calculate the absolute pressure at the bottom of the tank

Using formula of absolute pressure

$P=P_{atm}+\rho gh$

Put the value into the formula

$P=1.013\times10^{5}+1000\times9.8\times0.5$

$P=1.06200\times10^{5}\ Pa$

Hence, (a). The density of the wood is $1479.48\times10^{2}\ Kg/m^3$

(b). The absolute pressure at the bottom of the tank is $1.06200\times10^{5}\ Pa$ .

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