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3 years ago

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ماهي قوة التركيز من العين عندما ينضر الكاءن من ٢٠ الى ٥٠٠ من عينه

1 answer:

Debora [2.8K]3 years ago

4 0

Answer:

What is the power of focus from the eye when a subject looks from 20 to 500 from its eye?

Explanation:

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Pls help pls due today

BaLLatris [955]

The mutualism I believe. So sorry if I’m wrong

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3 years ago

Read 2 more answers

Tammy is a forensic investigator examining a body at a crime scene. She notes that the body is stiff but still flexible. What do

wlad13 [49]

I may be wrong but does it mean it was revent? because i know shortly after someone dies your body becomes fully stiff so maybe it was recent and it's in the process off stiffening up

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3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

3 years ago

The black hole is___<br> times smaller that the star.<br> I need answers please

mojhsa [17]

Answer:

The answer is 24 (for the first question).

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>

3 0

2 years ago

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, $a=1.8\ m/s^2$

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

At t = 1 s

$s=\dfrac{1}{2}\times 1.8\times 1^2$

$s_1=0.9\ m$

At t = 2 s

$s=\dfrac{1}{2}\times 1.8\times 2^2$

$s_2=3.6\ m$

At t = 3 s

$s=\dfrac{1}{2}\times 1.8\times 3^2$

$s_3=8.1\ m$

Hence, this is the required solution.

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3 years ago