It is to highly priced for most of civilization.
The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.
The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>
The answer using the graphical method and analytical method of vector addition will always be
C. Same
Analytic method means adding vectors (x₁,y₁) and (x₂,y₂) give (x₁+x₂,y₁+y₂)
Example: Addition of (2,3) and (1,1) gives (3,4)
Solving it graphically will also give (3,4)
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
