A 10kg object is 15 meters up a hill. Find its potential energy
1 answer:
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Answer:
Third displacement = 2.81 m which is 61.70° 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.
She sails 2 km east, displacement = 2 i
Then 3.50 km southeast, means 3.5 km 315⁰ to + ve X axis
Displacement = 3.5 cos 315 i + 3.5 sin 315 = 2.47 i - 2.47 j
Let third displacement be x i + y j
We have final displacement = 5.80 km east = 5.80 i
From summation we have total displacement = 2 i + 2.47 i - 2.47 j + x i + y j
= (4.47+x) i + (y - 2.47) j
Comparing both , we have 4.47+x = 5.80
x = 1.33
y-2.47=0
y = 2.47 j
So third displacement = 1.33 i + 2.47 j
Magnitude of third displacement =
θ = tan⁻¹(2.47/1.33) = 61.70°
So third displacement = 2.81 m which is 61.70° north of east.
