Answer:
Electric field E = kQ/r^2
Distance between charges = 6.30 - (-4.40) = 10.70m
Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.
Field from q1 at P = k(-9.50x^10^-6) / d^2
Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2
These fields are in opposite directions and are equal magnitudes if the resultant field = 0
k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2
9.50 / d^2 =8.40 / (10.70-d)^2
d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131
d/(10.70-d) = sqrt(1.1331) = 1.063
d = 1.063 ((10.70-d)
= 10.63 - 1.063d
2.063d = 10.63
d = 5.15m
The y coordinate where field is zero is 6.30 - 5.15 = 1.15m
Explanation:
Answer:
Oh I am sorry this is my first time on brainly i dont how to exit and sorry but dont know the answer
Explanation:
The 3rd Quadrant is the lower left one on a graph,
where ' x ' and ' y ' are both negative.
Your picture is completely unrelated to this question.
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s
