Calculate the total pressure at the bottom of a lake of depth 5.1 m (atm pressure `=10^(5)` P), density of water `=10^(3)kg//m^(3),g=10m//s^(2)`

If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid ( a_(y) ) for calculation of pressure, effective g is used. A closed box horizontal base 6 m by 6 m and a height 2m is half filled with liquid. It is given a constant horizontal acceleration g//2 and vertical downward acceleration g//2 . Water pressure at the bottomof centre of the box is equal to (atmospheric pressure =10^(5)N//m^(2) density of water =1000 kg//m^(3),g=10m//s^(2) )

A vessel full of water has a bottom of area 20 cm^(2) , top of area 20 cm^(2) , height 20 cm and volume half a litre as shown in figure. (a) find the force exerted by the water on the bottom. (b) considering the equilibrium of the water, find the resultant force exerted by the sides of the glass vessel on the water. atmospheric pressure =1.0xx10^(5) N//m^(2) . density of water =1000 kg//m^(3) and g=10m//s^(2)

What is the pressure at the bottom of the ocean at a place where it is 3 km deep ? [atmospheric pressure =1.01 xx10^(5)Pa, Density of sea water =1030kg m ^(-30)]

A body is experiencing an atmospheric pressure of 1.01xx10^(5)N//m^(2) on the surface of a pond. The body will experience double of the previous pressure if it is brought to a depth of (given density of pond water =1.03xx10^(3)kg//m^(2) , g=10m//s^(2) )

An experimentally decided formula for relation between pressure and density of x a liquid is (P)/(P_("atm"))=(B+1)((d)/(d_("atm")))^(n)-B B and n are constant. P_("atm") is atmospheric pressure and d_("atm") is the density of liquid at that pressure. For water numerical values of B=3000 and n=7. Take atmospheric pressure=100kPa and density of water =10^(3) kg//m^(3) at atmospheric pressure. At what depth is the density of water increase by 0.001%.

The pressure acting on a submarine is 3 xx10^(5) at a certain depth. If the depth is doubled the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is 1 xx 10 ^(5) Pa density of water is 10^(3) kgm ^(-3), g - 10 ms ^(-2))

The U-tube acts as a water siphon. The bend in the tube is 1m above the water surface. The tube outlet is 7m below the water surface. The water issues from the bottom of the siphon as a free jet at atmospheric pressure. Determine the speed of the free jet and the minimum absolute pressure of the water in the bend. Given atmospheric pressure =1.01xx 10^(5)N//m^(2),g=9.8 m//s^(2) and density of water =10^(3)kg//m^(3) .

A glass full of water has a bottom of area 20 cm^(2) , top of area 20 cm , height 20 cm and volume half a litre. a. Find the force exerted by the water on the bottom. b. Considering the equilibrium of the v , after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure = 1.0 xx 10^(5) N//m^(2) . Density of water = 1000 kg//m^(-3) and g = 10 m//s^(2)

The pressure at the bottom of a lake due to water is 4.9xx 10^(6)N//m^(2) . What is the depth of the lake?