Which type of star has surface temperatures bewteen 2400 and 50000 k
1 answer:
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Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = ( - v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²