physics help PLEASE!! 10 points..?

Question

A loaded flatbottom barge floats in fresh water. the bottom of the barge is 4.53 m below the water line. When the barge is empty the barge's bottom is only 1.08 m below the water line.
What is the difference between the pressure on the bottom of the loaded barge and the pressure at the water line?

I GOT THE FIRST PART.
the answer is 44394 Pa,
im not sure if its significant in answering the next part..


If the surface area of the bottom of the barge is 496 m^2, what is the weight of the load in the barge?

Answer in units of N.

like i said for the previous question, you dont have to give me the answer, a simple equation and explanation will be very helpful. THANKS!!

Answer 1

Hydrostatic Pressure in a Liquid

a)The pressure due to the liquid alone (i.e. the gauge pressure) at a given depth depends only upon the density of the liquid ρ and the distance below the surface of the liquid h.

ρ = 1000 kg / m^3
h = 4.53 m

ΔP = ρ g h = 44439.3 kg / (m s^2)

ΔP = 44439.3 Pa pressure difference

b)
In physics, buoyancy is the upward force on an object produced by the surrounding fluid (i.e., a liquid or a gas) in which it is fully or partially immersed, due to the pressure difference of the fluid between the top and bottom of the object. The net upward buoyancy force is equal to the magnitude of the weight of fluid displaced by the body.

The buoyant force can be expressed using the following equation:

Fbuoyant= ρwater V g

ρwater is the density of the water
V is the volume of the object submerged
g is the standard gravity ( 9.81 N/kg on Earth)


S = 496 m^2
D = 4.53 - 1.08 = 3.45 m
now when you put the load in the barge the barge goes in the water with a volume V = S*D = 1711.2 m^3

Fbuoyant= ρwater V g = G

G -weight of the load

G = 1000 *1711.2 * 9.81 = 16786872 N

Answer 2

Let
p= 1000 kg/m^3
g= 9.81m/s^2
a) The pressure below the water line is the column height of the water.
P= gph
P= 9.81 x 1000 x 4.53 =44400 Pa (3SF max)

However the difference in pressure between loaded and empty is

P= gp(h2-h1)
P=9.81 x 1000 (4.53 - 1.08)
P=33,800 Pa (3 SF)

b) The weight of the load is proportional to the additional volume displaced by submerging from h1 to h2.
Volume is V= Ah then
weight is
W= gpV2
W= gpA(h2-h1)
W=9.81 x 1000 x 496 (4.53 - 1.08 )=
W=16.8 E6 N (3SF)

(You can ignore the numerical answers given if you like :-) )

SF - least significant figure

Answer 3

i wont supply you an answer yet i will permit you comprehend the shape to do it a million. seperate the horizontal and vertical aspects which supply you the vertical and horizontal velocity (use trigonometry) 2. locate the displacement (distance) vertically tht it travels, and horizontally 3. keep in mind that horizontally there will be 0 acceleration except they supply you the air resistance, and vertically the acceleration would be 9.8N (by way of gravity) 4. additionally be sure you choose for on a favorable and destructive path (i anticipate which you will comprehend what im talking approximately) 5. hint: use the equations of action to locate the displacement (i additionally anticipate you're attentive to those) 6. the horizontal displacement may be the x fee, vertical would be y 7.hint: keep in mind which you in basic terms have those values to paintings with: the preliminary velocity (vertical and horizontal), the time (8 seconds), and the acceleration...they're the only values you like desire that facilitates ;)