A bowling ball collides with a pin, knocking it over. The ball continues to move
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Answer:
Explanation:
a ) angular frequency ω =
k is spring constant and m is mass attached
ω =
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2
The time when the cheetah catches up with the widebeest is 1.53 s
If the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.
The given parameters;
- speed of the wildebeest calf, Vw = 10 m/s
- distance traveled by the calf before the cheetah stands up = 7 m
- constant acceleration of the cheetah, a = 9.5 m/s²
Let the speed of the cheetah = Vc
let the time the cheetah catches up with the wildebeest = t
Apply relative velocity formula to determine the time when the cheetah catches up with the widebeest;
Assuming the wildebeest and the cheetah are running in the same direction;
The time when the cheetah catches up with the widebeest is 1.53 s
If the initial separation distance approaches zero;
Thus, if the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.
Lear more here: brainly.com/question/24430414
Answer:
Explanation:
t=Time airborne=2.54 s
R= horizontal range= 25.5 m
From the projectile motion equations we know that:
H=7.9 m