What amount of force is needed to propel and object of 27 kg to an acceleration of 11,550 m/s^2?
1 answer:
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Answer:
Explanation:
Since the cable touches the road at the mid point of two towers
so here we have vertex at that mid point taken to be origin
now the maximum height on the either side is given as
horizontal distance of the tower from mid point is given as
now from the equation of parabola we have
now we have
now we need to find the height at distance of 200 ft from center
so we have
Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
=
(
-
) ........................................................ i
And,
v = v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
=
(
-
)
or, = (
)
=(
)
or, y = 3.56 cm.