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3 years ago

Use the periodic table from the lesson to answer the following question. The atomic mass of F is ___. 9 10 19 55

Physics

2 answers:

Basile [38]3 years ago

7 0

18.9 but round it and you get 19

almond37 [142]3 years ago

4 0

<h2><u>What is the atomic mass of the element F?</u></h2><h3>A. 9</h3><h3>B. 10</h3><h3>C. 19</h3><h3>D. 55</h3><h3><u>Answer</u></h3>

In this case if given those choices your answer would be (C. 19).

However in more complicated Chemistry rules you would learn that the periodic table only showcases 1 of the many possible masses of the element.

Hope this help! :D

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A) what is the light source for a plant growing in the shade?

Ulleksa [173]

a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.

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3 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

You drop a ball from a height of 32 meters how much time passes before the ball hits the ground

tankabanditka [31]

Answer:

31 seconds

Explanation:

ajsfhlAS

5 0

3 years ago

What is the chemical name for a sand

vovikov84 [41]

It is known as silicon dioxide or silica!

Hope this helps!

7 0

3 years ago

A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s

Kryger [21]

The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

<u>Explanation:</u>

Given,

Temperature, T = 0°C

Initial mass, Mi = 62kg

Speed, s = 5.48m/s

Distance, x = 26.8m

Friction is present.

Mass of ice melted = ?

We know,

The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice

and

$Kinetic Energy, KE = \frac{1}{2} mv^2$

Therefore, $KE = \frac{62 X 5.48 X 5.48}{2}$

KE = 930.94 Joules

Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.

Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.

Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

4 0

3 years ago