2) A Ship has an area of cross - section at the water line of 2000m² (a) By what deck does the ship sink in fresh.
water, when it loads a cargo of 4000 tonnes
(b) if the ship + Cargo has a displacement tonnage
of 12300 tonnes; by what amount will the ship
rise in the water when it sails from fresh water
into Seawater (density of Sea water - 1025kgm-²​

a. 2 m

b. 0.15 m

Explanation:

(a) By what deck does the ship sink in fresh.

water, when it loads a cargo of 4000 tonnes

We know that the upward force , U on the ship equal the weight of fresh water displaced, W = ρVg where ρ = density of fresh water = 1000 kg/m³, V = volume of water displaced and g = acceleration due to gravity.

So, U = W = ρVg

This upward force must equal the weight of the ship, W' = mg where m = mass of ship = 4000 tonnes = 4000 × 1000 kg = 4 × 10⁶ kg

So. U = W'

ρVg = mg

V = m/ρ = 4 × 10⁶ kg/10³ kg/m³ = 4 × 10³ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in water h.

So, V = Ah

Thus h = V/A = 4 × 10³ m³/2000 m³ = 2 m

So, the ship sinks to a depth of 2 m in fresh water.

(b) if the ship + Cargo has a displacement tonnage

of 12300 tonnes; by what amount will the ship

rise in the water when it sails from fresh water

into Seawater (density of Sea water - 1025kgm⁻³​

We know that the upward force , U' on the ship in fresh water equals the weight of fresh water displaced, W" = ρV'g where ρ = density of fresh water = 1000 kg/m³, V' = volume of water displaced and g = acceleration due to gravity.

So, U = W" = ρV'g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U' = W₁

ρV'g = m₁g

V' = m₁/ρ = 1.23 × 10⁷ kg/10³ kg/m³ = 1.23 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in fresh water h'.

So, V' = Ah'

Thus h = V'/A = 1.23 × 10⁴ m³/2000 m³ = 6.15 m

So, the ship sinks to a depth of 0.6 m in fresh water.

Also,

We know that the upward force , U" on the ship in sea water equals the weight of sea water displaced, W₂ = ρ'V"g where ρ' = density of sea water = 1025 kg/m³, V"= volume of water displaced and g = acceleration due to gravity.

So, U" = W₂ = ρ'V"g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U" = W₁

ρ'V"g = m₁g

V" = m₁/ρ' = 1.23 × 10⁷ kg/1.025 × 10³ kg/m³ = 1.2 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in sea water h".

So, V" = Ah"

Thus h" = V'/A = 1.2 × 10⁴ m³/2000 m³ = 6 m

So, the ship sinks to a depth of 6 m in fresh water.

So, the rise in height from fresh water to sea water is Δh = h' - h" = 6.15 m - 6 m = 0.15 m

technology is the collection of techniques, skills, methods, and processes used in the the merriam-webster learner's dictionary offers a definition of the term: "the use of moreover, these technologies have become so complex that entire fields more recently, modern works of science fiction such as those by philip k.

explanation:

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂