I had this question already I still have my answer the answer is D is the correct answer.
Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
Answer:
There are different ways to investigate density. In this required practical activity, it is important to:
record the mass accurately
measure and observe the mass and the volume of the different objects
use appropriate apparatus and methods to measure volume and mass and use that to investigate density
Explanation:
Answer:
65
Explanation:
The resonant frequencies for a fixed string is given by the formula nv/(2L).
Where n is the multiple
.
v is speed in m/s
.
The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn
fundamental frequency means n=1
i.e fn+1 – fn = 390 -325
= 65
