What is the energy (in mev) released in the alpha decay of 239pu?
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Answer:
the magnitude of the dog's total displacement to the nearest tenths = 53.12°
Explanation:
Given that:
a dog walks 8.0m due to the north and 6.0 m due east.
We can determine the magnitude of the dog's total displacement by using the Pythagoras theorem.
Consider the north direction be the opposite and the east direction to be the adjacent, therefore the hypotenuse will be the dog's total displacement.
So, suppose hypotenuse is represented by a
the opposite represented by b, and the adjacent represented by c
Then;
a² = b² + c²
a² = 8² + 6²
a² = 64 + 46
a² = 100
a =
a = 10.0 m
However, the magnitude of the dog's total displacement is calculated by taking the tangent of the system.
tan θ = opposite / adjacent
tan θ = b/c
tan θ = 8/6
tan θ = 4/3
tan θ = 1.333
θ = tan⁻¹ (1.333)
θ = 53.12
the magnitude of the dog's total displacement to the nearest tenths = 53.12°