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Natali [406]
2 years ago
13

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?
Physics
1 answer:
astra-53 [7]2 years ago
6 0

Answer:

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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Explanation:

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due to this constant voltage difference the current in this conductor will remain constant

So here since the voltage across the conductor is constant so we can say that current in the conductor will remain constant and hence this is known as DC current.

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The only scalar quantity is a. 35 m

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Learn more about scalars and vectors:

brainly.com/question/2678571

brainly.com/question/4945130

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