Answer:
α = -π/3 rad/s²
θ = 1.5π rad ≈ 4.71 rad
θ = 0.75 rev
Explanation:
30 rev/min (2π rad/rev) / (60 s/min) = π rad/s
α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²
θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad
θ = 1.5π rad / 2π rad/rev = 0.75 rev
Answer:112.82 m/s
Explanation:
Given
range of arrow=68 m

as the arrow travels it acquire a vertical velocity 

-------1
Range is given by
R=ut
where u=initial velocity


substitute the value of t in eqn 1

--------2
and 

substitute it in 2


u=112.82 m/s
Answer:
Physics
Explanation:
Explanation:
We can use the Theorem of Work (W) and Kinetic Energy (K):
W=ΔK=Kf−Ki
it basically tells us that the work done on our system will show up as change in Kinetic Energy:
We know that the initial Kinetic Energy, Ki=12mv2i, is zero (starting from rest) while the final will be equal to 352J; Work will be force time displacement. so we get:
F⋅d=Ff
45d=352
and so:
d=35245=7.8≈8m
Heat can travel as<em> radiation</em>, even if there are no particles of matter
along the way. Which is really lucky, because that's how we get
heat from the sun. And light too.