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Nadusha1986 [10]
2 years ago
15

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

ing? Neglect friction and assume the ramp is plenty long enough to do this. (For this problem your initial kinetic energy can be set equal to your final potential energy. 1/2mv2 = mgh).
Physics
2 answers:
Zepler [3.9K]2 years ago
8 0

Answer:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

vivado [14]2 years ago
5 0

Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2             KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m

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Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

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A)

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B)

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