PHYSICS HELP:What is the distance from the top of the block to the water if the water is fresh?

PHYSICS HELP: What is the distance from the top of the block to the water if the water is fresh?

A 10cm by 10 cm by 10 cm wood block with a density of 700kg/{m}^{3} floats in water.

a.What is the distance from the top of the block to the water if the water is fresh?

b.If it’s seawater?

Answer:

(a).

density of fresh water is ρ_water = 1000 kg/m³

ΣF = 0

– (ρ_wood)g(V_wood) + (ρ_water)g(v_wood) = 0

(v_wood)/(V_wood) = (ρ_wood)/(ρ_water)

(10 * 10 * h)/(10 * 10 * 10) = (700)/(1000)

h = 7 cm

the distance from the top of the block to the water is 10 – 7 = 3 cm

(b).

density of seawater is ρ_sea = 1030 kg/m³

(v_wood)/(V_wood) = (ρ_wood)/(ρ_sea)

(10 * 10 * h’)/(10 * 10 * 10) = (700)/(1030)

h’ = 6.79611 cm

the distance from the top of the block to the water is 10 – 6.79611 = 3.20389 cm

if an object floats in water then the fraction of its volume that floats inside water is fiven by

d1/d1 …where d1=density of the floating object and d2=density of water

hence fraction of volume of wood that floats inside water is given by

700/1000 (density of water=100 kg/m^3)

=0.7 part of volume floats inside water

hence 0.3 part of of vulme floats outside water

now volume of wood=10 *10*10=1000 cm^3 =10^-3 cubic metre

hence volume of wood outside water=0.3 *10^-3= 3*10^-4 cubic metre

height of block from water * base area =volume outside water

h*10 ^-2 metre sq.=3*10^-4 metre cube

h=3* 10^-2 m= 3 cm…this is the answer

The wood is 70% of the weight of the water, so 70% of the wood will submerge leaving 30% floating above water line.

a. distance above water = 3 cm

——————————————–

b. Seawater weighs 1030 kg /m^3

700 / 1030 = 67.96 % as heavy as seawater

100% – 67.96% = 32.04 % of wood will be above waterline

.3204* 10cm = 3.2 cm of wood will be above waterline

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