So we want to know what do electric field lines show except strength of the electric field. Electric field lines are lines that show us the field strength and the DIRECTION of the electric field. If the lines point away from the charge that is producing the lines, the charge is positive, and if the lines point towards the charge that is producing them that charge is negative.
If both particles have the SAME electrical charge, then they repel.
If they have DIFFERENT electrical charge, then they attract.
Protons have + charge .
Electrons have - charge .
So two protons (A) or two electrons (D) push apart.
One proton and one electron (C) pull together.
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
Oxygen has<span> a higher electro negativity that then Sulfur, so Sulfur </span>will<span> " lose" electrons to Oxygen and that </span>is<span> the electrons </span>will be<span> pulled closer to the Oxygen causing, for oxygen to </span>have a negative<span> charge and the Sulfur to </span>have<span> a positive charge</span>
Explanation:
We have,
Speed of plane a is 900 km/h
Plane b is moving at a rate of 
It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.
Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.
