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Andrej [43]
3 years ago
9

The electron dot diagram for a neutral atom of chlorine (atomic number 17) is shown below. Which of the following symbols repres

ents a chlorine ion with a stable arrangement of eight valence electrons?
A. 35Cl1-
B. 35Cl2-
C. 35Cl1+
D. 35Cl


  .  .
: C :
   .
 
this is the symbol 
Physics
1 answer:
rosijanka [135]3 years ago
7 0
<span><span>The answer is A.

</span><span>Chlorine is found in group7A of the periodic table. This means that when they are neutral, (electrons and protons are equal), they lack 1 electron to complete the octet or eight valence electrons to be stable. This will result in Chlorine having to gain an electron. Since it gained an electron, that means the electrons outnumber the protons by 1, which will result in a negative charge of -1. </span></span>
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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
3 years ago
A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleratio
rjkz [21]

Hi there!

We can use Newton's Second Law:

\Sigma F = ma

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

4 0
2 years ago
Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle
Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

T = 1.812x10^{7} Km

\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU

T = 0.120 AU

T = \frac{0.120 AU}{1 AU} . 1 year

T = 0.120 year

T = \frac{0.120 year}{1 year} . 365.25 days  

T = 43.83 days

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (69000 Km)}{(43.83 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

43.83 days . \frac{86400 s}{1 day} ⇒ 3786912 s

69000 Km . \frac{1000 m}{ 1 Km} ⇒ 69000000 m

v = \frac{2 \pi (69000000 m)}{(3786912 s)}

v = 114.483 m/s

 

\frac{114.483 m/s}{81.778 m/s} = 1.399            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

7 0
3 years ago
A rocket started from the rest and goes straight up with an acceleration of 4.0 m/s^2 for the first 100 seconds. Then it coasts
Dimas [21]

Answer:

idk

Explanation:

3 0
3 years ago
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A copper bar of length h and electric resistance r slides with negligible friction on metal rails that have negligible electric
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