Suppose you pull two opposite sides of the loop away from each other, such that the other two sides get closer, reducing the loo
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Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
(a) The ball has a final velocity vector
with horizontal and vertical components, respectively,
The horizontal component of the ball's velocity is constant throughout its trajectory, so , and the horizontal distance <em>x</em> that it covers after time <em>t</em> is
It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:
The vertical component of the ball's velocity at time <em>t</em> is
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:
So, the initial velocity vector is
which carries an initial speed of
and direction <em>θ</em> such that
(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is
so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of