You can see what is the electron configuration by looking at the layout of the periodic tables. the first shell will have a max of 2 electrons on it, once the first one is filled up a second is added with a max of 8 electrons on it and so on with the 8 as a max. so He, and H will only have them on the first shell but every horizontal row is a new valence or outer shell. so lets say for carbon look at the number in the upper left corner of the box will tell you the total number of electrons you will need. so start off with the first two electrons on the first shell. now you know that carbon needs 6 electrons in total, since you can only have a max of 2 on the first shell you need a second one so on the second one you will have to have the remaining 4. now elements are most stable when they have a full valence shell becuase those are the only electrons that will react with others. so if carbon has 4 it wants to either gain or lose 4 electrons so you could say that it would bond with 4H since each H will donate 1 electron to the C valence shell making all the H and C stable. CH4(methane)
Answer:
The answer is C. Seismic waves.
Explanation:
- <u>Seismic waves are waves of energy that travel through Earth's layers, and are a result of earthquakes.</u>
Answer:
The correct option is (b).
Explanation:
We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.
We know that, the work done is equal to the kinetic energy of an object i.e.

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.
Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m