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koban [17]
3 years ago
6

Use your periodic table to answer the following question.

Physics
1 answer:
Pavel [41]3 years ago
6 0
Helium has only 2 electrons in it's valence shell

So, option D is your answer.

Hope this helps!
You might be interested in
MGS and later orbiters found spectral evidence for what minerals on the Martian surface? Check all that apply.
slamgirl [31]

Mars Global Surveyors (MGS) and later orbiters found the following  minerals on the  Martian surface;

  • Phyllosilicates
  • Carbonate
  • Sulfates
  • Iron oxide

The Mars Global Surveyors (MGS) and later orbiters suggest that the Martian crust contains a higher percentage of volatile elements such as Sulphur and chlorine than the Earth's crust does.

These scientists also conclude that the most abundant chemical elements  in the Martian crust are those found in Igneous rock.

These elements include the following;

  • Silicon,
  • Oxygen,
  • Iron,
  • Magnesium,
  • Aluminum,
  • Calcium, and
  • Potassium.

They also, suggest that hydrogen is found in ice (water) while carbon is found in carbon dioxide and carbonates.

From the given options the minerals found in Martian surface include;

  • Phyllosilicates ------ these are sheet of silicate minerals
  • Carbonate
  • Sulfates
  • iron oxide

Learn more here: brainly.com/question/20470323

6 0
2 years ago
A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t
andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be (  R+ h )

orbital velocity  V₀ =  \sqrt{\frac{GM}{( R+h)} }

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x \sqrt{\frac{( R+h)}{GM } }

\frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } = R +h

h = \frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } - R.

7 0
3 years ago
Read 2 more answers
A cannon fires a 0.652 kg shell with initial
laila [671]
 Mass have no effect for the projectile motion and  u want to know the  height "h"   
first,
        find the vertical and horizontal components of velocity 
 vertical component of velocity = 12 sin 61                  
horizontal component of velocity = 12 cos 61
now for the vertical motion ;               
             S = ut + (1/2) at^2
where
 s = h 
u = initial vertical component of velocity 
t = 0.473 s 
a = gravitational deceleration (-g) = -9.8 m/s^2    
       
         h=[12×sin 610×0.473]+[−9.8×(0.473)2] 

u can simplify this and u will get the answer

h=.5Gt2 

H=1.09m
6 0
3 years ago
Read 2 more answers
Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes tha
Dahasolnce [82]

Answer:

Sarah is right

Explanation:

This is an exercise that differentiates between scalars and vectors.

A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.

In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero

consequently Sarah is right

4 0
3 years ago
A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0
3 years ago
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