Ahockey puck travels 2.00 m at 10.0 east of south before ricocheting 2.50 m at 75.0° north of east. what is the puck’s resultant displacement?

 Magnitude of displacement = 1.09 m

 Direction of displacement= 23.75⁰ north of east.

Explanation:

 Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

 A hockey puck travels 2.00 m at 10.0° east of south, that is hockey puck travels 2.00 m at 280.0° from positive x-axis anti clock wise.

   Displacement = 2 cos 280 i + 2 sin 280 j = 0.35 i - 1.97 j

Then it travels 2.50 m at 75.0° north of east, that is hockey puck travels 2.50 m at 75.0° from positive x-axis anti clock wise.

   Displacement = 2.5 cos 75 i + 2.5 sin 75 j = 0.65 i + 2.41 j

 So, Total displacement = 0.35 i - 1.97 j + 0.65 i + 2.41 j = 1.00 i + 0.44 j

  Magnitude of displacement =

  Direction of displacement, θ = tan⁻¹(0.44/1.00) = 23.75⁰ = 23.75⁰ north of east.

what type of charge does it have? negative

the device is called “ rocket engine”