The pressure at a depth of 200 m below the surface of the sea, will be ___KPa.

The density of seawater is 1024 kg/m³.

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Solution

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by dgh.
The given values are: d = 1024 $\frac{k g}{m^{3}}$ , h = 200 m and g = 10 $\frac{m}{{s e c}^{2}}$ .
Thus, pressure = dgh = 1024 $\times$ 200 $\times$ 10 = 2048,000 Pa = 2048 KPa .

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The pressure at a depth of 200 m below the surface of the sea, will be ___KPa. The density of seawater is 1024 kg/m3.

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by dgh. The given values are: d = 1024 kgm3, h = 200 m and g = 10 msec2. Thus, pressure = dgh = 1024 × 200 × 10 = 2048,000 Pa = 2048 KPa .

The pressure at a depth of 200 m below the surface of the sea, will be ___KPa.

The density of seawater is 1024 kg/m³.

The expression of pressure at the bottom of a liquid of height ‘h’ and density ‘d’ is given by dgh.
The given values are: d = 1024 $\frac{k g}{m^{3}}$ , h = 200 m and g = 10 $\frac{m}{{s e c}^{2}}$ .
Thus, pressure = dgh = 1024 $\times$ 200 $\times$ 10 = 2048,000 Pa = 2048 KPa .