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:
1 N
Explanation:
Buoyant Force: This is also called upthrust, It can be defined as the force which act upward exerted by a fluid when an object is placed in it.
The S.I unit is Newton.
From the question,
Buoyant force = Weight of the object in air- weight of the object when submerged in water.
U = W-W'.......................... Equation 1
Where U = upthrust, W = weight in air, W' = weight when submerged in water.
Given: W = 3 N, W' = 2 N
Substitute into equation 1
U = 3-2
W = 1 N
Answer
given,
mass of ball = 5.93 kg
length of the string = 2.35 m
revolve with velocity of 4.75 m/s
acceleration due to gravity = 9.81 m/s²
T cos θ = mg
T cos θ = 
T cos θ = 58.17
T² - 56.93T - 3383.75 = 0
T = 93.22 N
θ = 51.39°
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
