Phy Archimedes' Principle 問題求解?

[匿名]

2016-10-23 1:42 pm

1. A cubical block of wood of side 0.5 m floats in water. Given that the densities of wood and water are 800 kg/m^3 and 1000 kg/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. An object of mass 100 kg is tied to the bottom of a pond with a light chain. If the tension in the chain is 200 N, what is the mass of water displaced by the object ?

3. Determine the pressure at the bottom of an open tank if it contains layers of 10 cm of oil, 30 cm of water and 5 cm of mercury. Given that the densities of oil, water and mercury are 850 kg/m^3 , 1000 kg/m^3 and 13600 kg/m^3 respectively.

4. The density of a cubical metal box is 3500 kg/m^3 . If the side of the box is 1 m, determine its apparent weight when it is completely immersed in water of density 1000 kg/m^3.

回答 (1)

天同

2016-10-23 3:53 pm

✔ 最佳答案

1(a) Weight of block = 0.5^3 x 800g N = 981 N
(where g is the acceleration due to gravity, taken to be 9.81 m/s^2)

(b) By Principle of Flotation, weight of water displaced = 981 N

(c) Volume of water displaced = (981/9.81)/1000 m^3 = 0.1 m^3
Hence, depth of block immersed = 0.1/0.5^2 m = 0.4 m
-------------------
2. Upthrust = 100g + 200 = 1181 N
where g = 9.81 m/s^2, is the acceleration due to gravity
Hence, mass of water displaced = 1181/9.81 kg = 120.4 kg
-------------------
3. Pressure due to oil = (10/100) x 850g N = 85g N
where g = 9.81 m/s^2, is the acceleration due to gravity
Pressure due to water = (30/100) x 1000g N = 300g N
Pressure due to mercury = (5/100) x 13600g N = 680g

Hence, total pressure = (85 + 300 + 680)g N = 1065g N = 10,450 N
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4. Mass of box = 3500 x 1^3 kg = 3500 kg
Weight of box = 3500g N
where g = 9.81 m/s^2, is the acceleration due to gravity

Volume of water displaced = volume of box =1 m^3
Hence, upthrust = 1 x 1000g N
Therefore, apparent weight of box = (3500g - 1000g) N
= 2500g N = 24,525 N

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