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<math>h</math> = [[Depth]] in the [[fluid]]
===Calculating Pressure in a Fluid===
{| class="wikitable"
| style="height:20px; width:200px; text-align:center;" |A whale dives to a depth of 1300m below sea level. Given the density of [[water]] is 1000kg/m<sup>3</sup>, calculate the [[pressure]] on the whale due to the water correct to two [[Significant Figures|significant figures]].
| style="height:20px; width:200px; text-align:center;" |Normal atmospheric pressure at sea level is 101kPa. A miner works in a mine shaft 550m below the surface of the Earth. Given the density of the [[Earth's Atmosphere|atmosphere]] at this level is 1.2kg/m<sup>3</sup>, calculate the [[pressure]] on the miner correct to two [[Significant Figures|significant figures]].
| style="height:20px; width:200px; text-align:center;" |A piece of [[Lithium]] is submersed 4.3cm deep in oil prevent it from oxidising. The oil has a [[density]] of 932kg/m<sup>3</sup>, calculate the [[pressure]] on the [[Lithium]] due to the oil correct to two [[Significant Figures|significant figures]].
|-
| style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities'''
[[Density]] = 1000kg/m<sup>3</sup>
[[Gravitational Field Strength]] = 9.8N/kg
[[Depth]] = 1300
| style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities'''
[[Density]] = 1.2kg/m<sup>3</sup>
[[Gravitational Field Strength]] = 9.8N/kg
[[Depth]] = 550m
| style="height:20px; width:200px; text-align:left;" |'''1. State the known quantities'''
[[Density]] = 932kg/m<sup>3</sup>
[[Gravitational Field Strength]] = 9.8N/kg
[[Depth]] = 4.3cm = 0.043m
|-
| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
<math> P = \rho g h</math>
<math> P = 1000 \times 9.8 \times 1300</math>
<math> P = 12740000Pa</math>
<math> P \approx 13000000Pa</math>
| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
<math> P = \rho g h</math>
<math> P = 1.2 \times 9.8 \times 550</math>
<math> P = 6468Pa</math>
Since the [[miner]] also has normal atmospheric pressure 101kPa is added.
<math> P_{Total} = 6468 + 101000</math>
<math> P_{Total} = 107468Pa</math>
<math> P_{Total} \approx 110000Pa</math>
| style="height:20px; width:200px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
<math> P = \rho g h</math>
<math> P = 932 \times 9.8 \times 0.043</math>
<math> P = 392.7448Pa</math>
<math> P \approx 390Pa</math>
|}