Question

0/2 points Previous Answers SerCP11 15.2.OP.013. My Notes Ask Your Teacher The figure below shows three charged particles at the corners of an equilateral triangle. Particle A has a charge of 1.25 C; B has a charge of 5.60 C; and C has a charge of -5.04 C. Each side of the triangle is 0.500 m long. What are the magnitude and direction of the net electric force on A? (Enter the magnitude In N and the direction in degrees below the x-axis.) magnitude 0.0475 X Draw the directions of the forces acting on A, paying attention to the signs of the charges. From Coulomb's law, what are the magnitudes of these forces? What are their x and y-components? How can you use the components to find the net force? Be careful with signs and unit conversions. N 16.62 x Use there and y-components of the net force on A to find the angle. Note the angle is measured from below the positive x-axis (clockwise). below the x-axis direction Need Help? 2/2 points Previous Answers SerCP11 15.2.P.005. es which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles Chelium nuclel, each My Notes Ask Your Teacher MacBook Air

Answer 1

The figure shows three egnillatteral triangle. charged pastiles at the corner of an particle - A = 1.25MC 2 G has change particle B = 5.6 Me - – 5.04 o re Now the magnitude of the force on A charge 12 - Gx10x1.25x5 60x10 DN- = 9x10xlor 1.25x5 60x10 4156x (.5) so Tox (-25) Guct o fia der (25) Koi 5.04x 125 x 10 gx 18 x 125+ 5.0 4x70 F & 4THx(.5) - (4) 082852810 u fir , 252 Newton F = 2268 Newton Now The magnitude of the two forces. If I = √firt fh + 2 fif Cos (128) = 7602 Newton ..The magnitude of sthe force = .76 Newton . Now - 2268 = the angle taid = 252 id=tai (69) = 40