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BaLLatris [955]
3 years ago
13

At atom whose valence electron shell is nearly full is ________ chemically reactive. a. not b. mildly c. highly d. the name of t

he element is needed to figure this out.
Physics
2 answers:
svet-max [94.6K]3 years ago
5 0
I believe the correct response would be B. Mildly reactive. As the atom simply needs only a few more electrons in its valence shell to achieve stability and have an octet of 8 valence electrons in its valence shell.
Vaselesa [24]3 years ago
4 0
I bet is a because it will not be chemically
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Naily [24]
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"<span> if the shells are affected by the level of acidity, then they will be weaker.</span>"

That hypothesis refers directly to the theory being questions in the problem.

7 0
3 years ago
I dont know what it is so help me im depending you on this one with my grades
tigry1 [53]
“Newton's first law says that the tableware will remain motionless unless acted upon by an outside force. To set the objects on the table into motion, the horizontal force acting upon them in this case is the frictional force between them and the table cloth. “

couldn’t paraphrase i’m not too good at physics but this might help
3 0
3 years ago
A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partia
Vika [28.1K]

Answer:

a). Acceleration with old model a_{o}=8.343\frac{m}{s^{2}}

b). Acceleration with new model a_{n}=11.785\frac{m}{s^{2}}

c). Coefficient of kinetic friction old model u_{k}=0.0515

d). Coefficient of kinetic friction old model u_{k}=0.0464

Explanation:

Using the Newton laws so the initial speed is zero

so:

Δx=v_{o}*t+\frac{1}{2}*a*t^{2}

v_{o} because the ski began from the rest initial velocity is zero

Δx=\frac{1}{2}*a*t^{2}

a).

a=\frac{2*x}{t^{2}}

a_{o}=\frac{2*200m}{(47.94s)^{2}}

a_{o}=8.343\frac{m}{s^{2}}

b).

a=\frac{2*x}{t^{2}}

a_{n}=\frac{2*200m}{(33.94s)^{2}}

a_{n}=11.785\frac{m}{s^{2}}

In the horizontal direction the friction force acts on the ski is:

f_{k}=u_{k}*m*g*cos(\alpha)

u_{k}=\frac{m*g*sin(\alpha)-m*a}{m*g*cos(\alpha)}

c).

u_{k}=\frac{g*sin(\alpha)-a}{g*cos(\alpha)}

u_{k}=\frac{9.8*sin(4)-0.108}{9.8*cos(4)}

u_{k}=0.0515

d).

u_{k}=\frac{g*sin(\alpha)-a}{g*cos(\alpha)}

u_{k}=\frac{9.8*sin(4)-0.23}{9.8*cos(4)}

u_{k}=0.0464

5 0
3 years ago
What type of bond is shown in the figure?
ladessa [460]

Answer:

ionic

Explanation:

6 0
3 years ago
Read 2 more answers
If you accelerate from rest at 10 m/s^2 what will be your speed after 6s
trapecia [35]
Ok so we know:
The acceleration (10ms^-2)
The initial velocity (0)
And the total time (6seconds)

Using the equation v = u + at, with v being the final velocity, u being the initial velocity, a being acceleration and t being time, we get:

v = 0 + 10x6
v = 60

So the speed after 6 seconds is 60 metres per second (6m/s)

Hope this helped
5 0
3 years ago
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