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OleMash [197]
3 years ago
7

A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th

at the balance reads 4.3 N. What is the density of the block?
Physics
1 answer:
DiKsa [7]3 years ago
4 0

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

                        = 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= \frac{0.4937}{55\times10^{-6}}

Density of block = 8977.7 kg/m^3

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Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

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On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

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3 years ago
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ruslelena [56]

Answer:

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3 years ago
Any help guys? I am stuck on two problems.
Juli2301 [7.4K]

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8 0
3 years ago
Light is incident along the normal to face AB of a glass prism of refractive index 1.54. Find αmax, the largest value the angle
marusya05 [52]

To solve this problem it is necessary to use the concepts related to Snell's law.

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Therefore the \alpha_{max} would be equal to

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\alpha = 30.27\°

Therefore the largest value of the angle α is 30.27°

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3 years ago
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