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# Pressures problems are featured.

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1.) The water in the tank is at a guage pressure of 216,302 Pa If the bottom of the tank has an area of 0.6m^2 what is the force on the bottom of the tank.

2.) A Viewing window in an aquarium is 62 inches wide and 56 inches high. If the average pressure inside the water tank is 16psi. What is the outward force on the window.

3.) An object is recovered from under the sea. Its weight is measured to be 42.8lb and its volume 4 ft^3. What is the object's weight density?

4.) What is the guage pressure in psi at a depth of 30ft. in the ocean (salt water obviously). .433h guage pressure is only applicable in fresh water.
5.) what is the absolute pressure in PSI at the depth of 73ft in the ocean. Use 1 atm for the pressure in the atmosphere.

6.) A Juniper wood plank meausuring 8ft x 0.6ft x 0.1ft is totally emersed in water. what is the net force acting on the plank. (I know this formual Fnet=Fb-weight.

7.) Using Pascals Principle Given that F-out is 98lb The areas of the output cylinders are A-out 25in^2 and A-in 6in^2 what is the input force.

8.) A person with a weight of 235.6 lbs stands on two feet with there weigth equally distributed between the two shoes who's area is 39.7 square inches what is the pressure on each shoe independently.

9.) the total mass on a helium balloon is 127kg what is the volume of the ballon.

10.) An Empthy storage tank has a volume of 130ft^3. What is the boyuant force exerted on it by the air.

https://brainmass.com/physics/applied-physics/pressures-problems-are-featured-230209

#### Solution Preview

Hi Client,
Here are the solutions of these problems:

1)

Pressure = Force/Area
P = F/A
F = P.A
= 216302(N/m^2). 0.6(m^2)
= 129781 N

2)

F = P.A
= 16 (lb/in^2) . (62 x 56) (in^2)
= 55552 lb

3)

Density = weight/volume
= 42.8/4
= 10.7 lb/ft^3
4)

Pressure = hdg

h = 30 ft
= 30 x 12 inch
= 30 x 12 x 2.54 cm ... ( 1 in = 2.54 cm )
= 914.4 cm
= 9.14 m
P = hdg
= 9.14 x 1000 x 9.8 ..... ( density of sea water = 1000 Kg/m^3 approximately )
= 89572 Pa
= 89572 x 145 x 10^-6 psi ......... ( 1 Pa = 145 x 10^-6 psi )
= 12987940 x 10^-6
= 12.98 psi
= 13 psi

5)

P = hdg

h = 73 ft
= 73 x 12 in
= (73 x 12 ) x 2.54 ...

#### Solution Summary

The solution examines pressure problems in tanks and aquarium.

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