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aev [14]
3 years ago
5

Moving from the element with atomic number 10 to atomic number 11 on the periodic table, there is a change in reactivity. In thr

ee to five sentences, identify the direction of the change, and give two reasons for the change.(4 points)
Physics
1 answer:
bazaltina [42]3 years ago
7 0

Answer:

Explanation:

Ne has its outside shell full of electrons; therefore, its reaction with other elements is minimal if at all. Na, the next element, has 1 more proton and 1 more electron. More importantly that added electron is in the outside shell. It can fill that shell by ADDing 7 more electrons or LOSing the one electron. Obviously, it loses the one electron and it does so with gusto which makes Na a very reaction element while Ne is not reactive at all. Now follow oobleck's instructions, add what you read to what I've written,then summarize all of it in your own words.

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Water waves, earthquake waves, sound waves, and the waves that travel down a rope or spring are all examples of what waves.
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Gamma ray technology can be used to do which of the following?
Rudiy27

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3 years ago
Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l
Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0
3 years ago
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3 0
3 years ago
A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo
Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0
3 years ago
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