Question

A square plate of side 10 m is placed horizontally 1 m
below the surface of water. The atmospheric pressure
is 1.013 x 105 N m-2. Calculate the total thrust on the
plate.
(Density of water p = 103 kg m-3, g = 9.8 m s-2)

Answer only if u know ​

Answer 1

Answer:

$total \: thrust \: on \: the \: plate = 1.111 \times 10 {}^{7} N.$

Step-by-step explanation:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:here$

$atmospheric \: pressure \: \: \: P _{0} = 1.013 \times 10 {}^{5} N \: m {}^{ - 2}$

$density \: of \: water \: \: \: \: p = 10 {}^{3} kg \: m {}^{ - 3}$

$g = 9.8m \: s {}^{ - 2}$

$\: \: h = 1m$

$total \: pressure \: at \: a \: point \: 1m \: \\ below \: the \: surface \: of \: water \\ = P _{0}(atmospheric \: pressure) + \\ h \: p \: g(pressure \: due \: to \: a \\ \: column \: of \: water \: of \: height \: 1m)$

$= P _{0} + h \: p \: g$

$= (1.013 \times 10 {}^{5}) + (1 \times 10 {}^{3} \times 9.8)$

$= (101300) + (9800) \\ = 111100 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$therefore \: total \: pressure = 1.111 \times 10 {}^{5} N \: m {}^{ - 2}$

$area \: of \: plate \: = 10m \times 10m = 100m {}^{2}$

$therefore \: total \: thrust \: on \: the \: \\ plate = pressure \times area \: of \: \\ plate$

$that \: is \: = (1.111 \times 10 {}^{5} ) \times 10 {}^{2} \\ = 1.111 \times 10 {}^{7} N.$

Hope you find this useful ◖⚆ᴥ⚆◗