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zysi [14]
3 years ago
14

A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely i

mmersed in water, the scale reads 18. 2 N. What are the volume and density of the block?
Physics
2 answers:
blsea [12.9K]3 years ago
8 0

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

Schach [20]3 years ago
5 0

Answer:

The volume of the block is 306 cm³

The density of the block is 7.07 g/cm³

Explanation:

Given;

weight of block in air, W_a = 21.2 N

Weight of block in water, W_w = 18.2 N

Mass of the block in air;

W_a = mg

21.2 = m x 9.8

m = 21.2 / 9.8

m = 2.163 kg

mass of the block in water;

W_w = mg

18.2 = m x 9.8

m = 18.2 / 9.8

m = 1.857 kg

Apply Archimedes principle

Mass of object in air  - mass of object in water = density of water   x  volume                  of object

2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block

0.306 kg = 1000 kg/m³ x Volume of block

Volume of the block = \frac{0.306 \ kg}{1000 \ kg/m^3}

Volume of the block = 3.06 x 10⁻⁴ m³

Volume of the block = 306 cm³

Determine the density of the block

Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3

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Answer:

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Explanation:

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Explanation:

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Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

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Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

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