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egoroff_w [7]
3 years ago
11

A 6.0 kg bucket of water is raised from a well by a rope. if the upward acceleration of the bucket is 3.9 m/s2, find the force e

xerted by the rope on the bucket
Physics
1 answer:
shutvik [7]3 years ago
8 0
I'm not good with math but I think it is 23.4
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What are the equivalent celsius and kelvin temperatures of -128 f?
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Answer:

128 Kelvin = 128 - 273.15 = -145.15 Celsius. Temperature conversion chart Sample temperature conversions 103.55 Kelvin to degrees Fahrenheit 39.82 degrees Fahrenheit to Kelvin

Explanation:

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What is 7.832 rounded to the nearest 10th?
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When rounding you look at the number right after the one you want to round. The tenth place is the one right after the decimal. If the number after it is more than five it goes up by one, it it's less than five it stays the same. So since the number after the 8 is a 3 it stays the same.

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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.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

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

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

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

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