a rectangular tank 4 m long , 1.5m wide contains water upto a height of 2m . Calculate the force due to water pressure on the base of the tank . Find also the depth of the centre of pressure from the free surface.
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Tank is moving upwards with an acceleration of
Tank is moving upwards with an acceleration of 2g
Tank is moving upwards with an acceleration of 2g
Tank is moving upwards with an acceleration of 2g m/s
Tank is moving upwards with an acceleration of 2g m/s 2
Tank is moving upwards with an acceleration of 2g m/s 2
Tank is moving upwards with an acceleration of 2g m/s 2 Thus for the system g eef = 23g
3g
3g m/s
3g m/s 2
3g m/s 2
3g m/s 2 Pressure increases with depth according to the formula : P = P
3g m/s 2 Pressure increases with depth according to the formula : P = P 0
3g m/s 2 Pressure increases with depth according to the formula : P = P 0
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff h
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0 = 0. Because atmospheric pressure is applied from both sides thus it;s cancelled.
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0 = 0. Because atmospheric pressure is applied from both sides thus it;s cancelled.∴ Pressure = ρg
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0 = 0. Because atmospheric pressure is applied from both sides thus it;s cancelled.∴ Pressure = ρg eff
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0 = 0. Because atmospheric pressure is applied from both sides thus it;s cancelled.∴ Pressure = ρg eff
3g m/s 2 Pressure increases with depth according to the formula : P = P 0 +ρg eff hAlso pressure due to the weight of the liquid = ρVg/A = ρVh=22.5㎩Here P 0 = 0. Because atmospheric pressure is applied from both sides thus it;s cancelled.∴ Pressure = ρg eff h=18742.5㎩ + 22.5㎩ =187.65 kPa
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