A mass oscillating up and down on the bottom of a spring can be modeled as harmonic motion. if they weight is displaced a maximum of 7 cm and it takes 4.0 sec to complete one cycle to the nearest hundredth of a cm how far above or below the resting position is the weight after 1.15 seconds ?

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  1.63 cm (across the centerline from release)

Step-by-step explanation:

If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...

  h(t) = -7cos(2πt/4)

  h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm

That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.

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If it is released from above the resting position, it will be 1.63 cm below the resting position at t=1.15 seconds.

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.

you never say the total so we cant find the percentage used for petrol

step-by-step explanation: