Archimedes’ Principle

1.A person of mass 50kg floats in water almost completely immersed.

a)Find (i) the weight of the person, (ii) the upthrust acting on him and (iii) the

weight of water displaced by the person.

b)If he now floats in denser salt water, how does this affect (i) the upthrust,(ii) the weight of salt water displaced and (iii) the level to which he is immersed in the

water?

2.An average person has a volume of 0.05m^3. What is the upthrust from

the air on the person if the density of air is 1.2kg/m^3?

3.A block of steel of density 7800kg/m^3 has a volume of 0.5^3. Calculate:

a)Its weight,

b)Its apparent weight when completely immersed in water of density

1000kg/m^3

.

4.A cubical block of wood of side 0.5m floats in water. Given that the

densities of wood and water are 800 and 1000kg/m^3 respectively, calculate:

a)The weight of the block,

b)The weight of water displaced by the block,

c)The depth of the block immersed in water.

2 個解答

評分
  • 天同
    Lv 7
    6 年前
    最愛解答

    1. (a)(i) weight of person = 50g N

    where g is the acceleration due to gravity

    (ii) Upthrust = 50g N

    (iii) Weight of water displaced = upthrust = 50g N

    2. Upthrust = 1.2 x 0.05g N = 0.06g N

    3. (a) Weight = 7800 x (0.5^3)g N = 975g N

    (b) Upthrust = 0.5^3 x 1000g N = 125g N

    Apparent weight = (975 - 125)g N = 850g N

    4.(a) Weight of block = 800 x (0.5^3)g N = 100g N

    (b) Weight of water displaced = weight of block = 100g N

    (c) Depth of block immersed in water

    = 100g/[(0.5^2) x 1000g] m = 0.4 m = 40 cm

  • 6 年前

    Thank you very much!

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