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kow [346]
3 years ago
13

What other two things could be added to what you consume each day that could make you even healthier?

Physics
2 answers:
blondinia [14]3 years ago
4 0

Answer:

Carrots and apple (heathly and cleans and strengthens teeth)

To have a balenced diet you should have fiber and minerals, vitamins, fats, protines, carbohhydrates, and also water.

Two other things i know are good for you are:

Bluebarries they are rich in fiber, vitamins A and C they also boost up ur cardiovascular health

Beans are also great because they provide protein, fiber, and they are low in calories and free of saturated fat

Explanation:

Crisp fruits and raw vegetables, like apples, carrots and celery, help clean plaque from teeth and freshen breath.

Hope this helps <3 Need thanks? Comment /hearthelp

miv72 [106K]3 years ago
3 0

You could I corporate water into your diet to make yourself hydrated, for less teens are hydrating themselves nowadays. You can also eat more vegetables because vegetables contain nutrients that meats do not have, such as vitamin C.

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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

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