In BPC
tan\theta =a/b = 3/4
\theta = tan^-1(0.75)
\theta = 36.87 deg
BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m
Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C
Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C
Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C
Net electric field along X-direction is given as
Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C
Net electric field along X-direction is given as
Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C
Net electric field is given as
E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
Answer:
If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).
Explanation:
In the middle, when the object was changing position at a constant velocity, the acceleration was 0. This is because the object is no longer changing its velocity and is moving at a constant rate.
Answer:
A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same
B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same
Explanation:
A vehicle crash problem must be solved with the equation of the moment,
Initial instant Before crash
p₀ = m v₁ + mv₂
After the crash
= m + m
p₀ =
If the speed ratio before and after the crash is one
p₀ / = 1
We can assume that initially one of the cars was stopped
m v₁₀ = m
v₁₀ =
For the two speeds to be equal, the masses of the vehicles must be the same.
A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same
B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same