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alexandr1967 [171]
3 years ago
9

Guys can you text here please i have way to many textes on the other one and also remember was told to paste this to other comme

nt sections sooo...HEY PLS DON'T JOIN THEMEETING CALL OF A PERSON WHO'S ID IS 825 338 1513 (I'M NOT SAYING THE PASSWORD) HE IS A CHILD PREDATOR AND A PERV. HE HAS LOTS OF ACCOUNTS ON BRAINLY BUT HIS ZOOM NAME IS MYSTERIOUS MEN.. HE ASKS FOR GIRLS TO SHOW THEIR BODIES AND -------- PLEASE REPORT HIM IF YOU SEE A QUESTION LIKE THAT. WE NEED TO TAKE HIM DOWN!!! )
(Its girls to i hade one)i also did it but ileft when she said it i was like oh no what just happend lol !)





(if you guys give me his account you can get 100 points )
Physics
1 answer:
Nadusha1986 [10]3 years ago
6 0

Answer:

ok

Explanation:

mark me as brainlist

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Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0
3 years ago
How do very high density objects appear in an ultrasound?
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<span>Density is entirely unrelated to an object's size. It is a property of a given</span>
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3 years ago
a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin
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Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

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a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

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According to a college survey, 22% of all students work full time. find the mean for the number of 3) students who work full tim
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E[x] = np = (16)(0.22) = 3.52 
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Help ASAP <br><br>Identify the cause of foliation in rocks?
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