Question

A vessel contains water to a height 1.5 m. Taking the density of water 10^3 kg m^-3,acceleration due to gravity 9.8 ms^-2 and area of base of vessel 100cm^2, calculate:(a)pressure and (b)the thrust at the base of the vessel.

Answer 1

thrust at the base of vessel= hdg (h = height , d= density, g= gravity). =1.5×10^3×9.8
=14700 N
pressure= thrust/ area = 14700÷ (100×10^-4) = 1.47×10^6 Pa

Answer 2

$\bold{(a) \: Pressure} \\ \bold{Height = 1.5m} \\ \bold{Density = 1000 kg/m³} \\ \bold{Acceleration \: due \: to \: gravity \: = \: 9.8m/s²} \\ \bold \green{Pressure \: = \: hpg} \\ \bold{ = \: 1.5 \times 1000 \times 9.8} \\ \bold{ = \: 14700 Pa} \\ \bold \red{ = \: 1.47 \times {10}^{4} N/m² \: \: ....ans}$

$\bold{(b) \: Thrust \: at \: the \: base \: of \: vessel} \\ \bold{Pressure \: = \: \frac{force}{area} } \\ \bold \green{force \: = \: Pressure \times area} \\ \bold{ = \: 1.47 \times {10}^{4} \times 100} \\ \bold \red{ = \: 147 N \: \: ....ans}$

Hope it helps you...

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