3) If a picture frame were supported by a pair
of vertical wires, then the tension in each
wire would be half of the weight of the frame.
The frame is supported by wires positioned
at the angles shown below. If the frame
weighs 28.0 N, what is the tension in each
angled wire?
60
50.00

Ithink it’s 98 because jajajjajansmsmjs

Part a:   using momentum conservation we have:   direction x:   (15) (7.0) + 0 = (15) (4.2) cos (-20 °) + (10) v₂ (f) cosθ  (10) v₂ (f) cosθ = (15) (7.0) - (15) (4.2) cos (-20 °) (eq 1)  direction y:   0 + 0 = (15) (4.2) sin (-20 °) + (10) v₂ (f) sinθ  (10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °) (eq 2)  we divide both equations (1) and (2):   left side:   (10) v₂ (f) sinθ / (10) v₂ (f) cosθ = tanθ  right side:   - (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °)  clearing the angle we have:   θ = tan⁻¹ [- (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °)]  θ = 25 °  we use any of the equations to find the speed.  from (2) we have:   (10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °)  v₂ = - (15) (4.2) sin (-20 °) / (10) sin25 °  answer:   v₂ = 5.1m / s  note:     θ = 25 ° (direction is in the first quadrant, north of east)  part b:   initial kinetic energy:   k (i) = 0.5 (15) (7.0) ²  = 370j (rounded)  final kinetic energy:   ke (f) = 0.5 (15kg) (4.2m / s) ² + 0.5 (10kg) (5.1m / s) ²  = 260j (rounded)  answer:     a difference of 110j, so it was an inelastic collision.