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zhenek [66]
3 years ago
12

On an amusement park ride passengers accelerate straight downward from rest to 22.9 m/s in 2.2 s.

Physics
2 answers:
Kamila [148]3 years ago
8 0

Given,  

Initial Velocity (u) = 0 m/s  

Final Velocity (v) = 25.9 m/s  

Time (t) = 2.5 sec  

avg. acceleration = ?  

avg. acceleration = v-u/t  

avg = 25.9/2.5 = 10.36 m/s²  

avg = 10.36 m/s²

Sindrei [870]3 years ago
5 0

Answer:

Acceleration, a=10.4\ m/s^2

Explanation:

Given that,

Initial velocity of the ride, u = 0 (at rest)

Final speed of the ride, v = 22.9 m/s

Time taken, t = 2.2 s

Let a is the acceleration of the passengers on this ride. It can be calculated using the formula of the acceleration as :

a=\dfrac{v-u}{t}

a=\dfrac{22.9-0}{2.2}

a=10.4\ m/s^2

So, the time required to bring the car to a stop is 10.4\ m/s^2. Hence, this is the required solution.

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Akimi4 [234]

Answer:

E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

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In this case y = E

a) E = A sin Bt

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b) E = A cos (Bt + c)

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We substitute in the wave equation

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       v = C / B

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