If the separation distance is doubled, then the electric field decreases by a factor of 4.
<h3>What is the electric field strength?</h3>
We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.
We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.
Thus we know that;
E = Kq/r^2
Where;
E = electric field strength
q = magnitude of charge
r = distance of separation
Now;
E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2
E = 4 N/C
Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.
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Answer:
4. The equilibrium will shift to favor formation of NO2(g)
Explanation:
According to La Chatalier's Principle which states that when an equilibrium system undergoes changes either in temperature, volume or concentration; there will be in a change in the system in order to reach equilibrium.
From the above equation,
N2O4(g) ⇀↽ 2 NO2(g)
From the above reaction, there are 2 moles of gaseous product on the left and 1 mole of gaseous reactant.
Therefore, there are more moles of gases in the left hand side than the right hand side.
Because a decrease in volume favors the direction that produces fewer moles, an increase in volume will therefore shift this system towards the side with more moles of gases that is, more products are formed hence, this system will shift to right and produce more moles of products i.e more NO2(g) formed.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
An electron has a negative charge. Hope this helps.
Answer:
Explanation:
Given data
time=0.530 h
Average velocity Vavg=19.0 km/s
To find
Displacement Δx
Solution
The Formula for average velocity is given as
