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Ivahew [28]
3 years ago
9

A platinum sphere with radius 1.21 cm is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting on

the sphere, and the sphere's apparent weight. The densities of platinum and mercury are 2.14 × 104 kg/m3 and 1.36 × 104 kg/m3, respectively.
Physics
1 answer:
TEA [102]3 years ago
5 0

Answer:

1.556 N , 0.989 N, 0.567 N

Explanation:

Radius of sphere, r = 1.21 cm = 0.0121 m

density of platinum , d = 2.14 x 10^4 kg/m^3

density of mercury, d' = 13.6 x 10^3 kg/m^3

Volume of sphere, \frac{4}{3}\pi \times r^{3}= \frac{4}{3}\times 1.34\times \left (0.0121  \right )^{3}

V = 7.42 x 10^-6 m^3

Weight of sphere = volume of sphere x density of platinum x gravity

W = V x d x g = 7.42 x 10^-6 x 2.14 x 10^4 x 9.8 = 1.556 N

Buoyant force, B = Volume x density of mercury x gravity

B = 7.42 x 10^-6 x 13.6 x 10^3 x 9.8 = 0.989 N

Apparent weight = True weight - Buoyant force

Apparent weight = 1.556 - 0.989 = 0.567 N

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6 0
2 years ago
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest
shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

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Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

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Therefore;

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object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0
3 years ago
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