A cube of wood having an edge dimension of 18.0 cm and a density of 651 kg/m3 floats on water.(a) What is the distance from the horizontaltop surface of the cube to the water level?(b) How much lead weight has to be placed on top of the cube sothat its top is just level with the water?

(a) The distance will be "6.282 cm".

(b) Mass will be "2.03 kg".

(a)

From Archimedes' principle, we get

→

             

                 

                 

So,

The distance from horizontal top to the water level will be:

=

=

(b)

After placing the lead block of weight , we have

→

                     

                             

                             

Mass,

m = 2.03 kg  

Thus the above answer is right.

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A. 6.282

B. 2.03kg

Explanation:

A.

We solve using archimedes principle

L³pwood = L²dwater

We make d subject of the formula

d = Lpwood/pester

= 18x651/1000

= 18x0.651

= 11.718cm

Distance from horizontal top to water level

= 18-11.718

= 6.282cm

B.

When we place lead block

WL + L³pwoodg = L³pwaterg

WL = L³g(Pwater-Pwood)

= 0.18³x9.8(1000-651)

= 19.94N

19.94/9.8

= 2.03kg

The mass m is therefore 2.03kg

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