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iren2701 [21]
3 years ago
7

Question 1 of 15

Physics
1 answer:
stepan [7]3 years ago
8 0

Answer:

significant

Explanation:

The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).

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18. The orbits of planets being elliptical was one the planetary laws developed by
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It was a man named <span>Johannes Kepler. </span>
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With speeds up to 90 miles per hour, what is the fastest sport on ice?.
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Luge

Explanation:

8 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.

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

5 0
3 years ago
2. An alternating current is represented by the equation I=20sin 100mt.
Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

I=I_o sin\omega t

Where

I₀ is the peak value of current

\omega is angular frequency

As\ \omega=2\pi f

So,

f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz

We know that,

I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0
2 years ago
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