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.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~