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Korolek [52]
3 years ago
8

I'm stuck on question 3. Could anyone please explain it?

Physics
1 answer:
sergiy2304 [10]3 years ago
5 0
Peak voltage is 2
period is 40ms
frequency = 1/period = 25Hz
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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel
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Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
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Answer: ymax = 10084.2m
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The most interpersonal constructive passion response to relational conflict is..
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a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the
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What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a
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E = mgh +  \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh +  \frac{1}2} m  r ^{2}   \omega ^{2}  + \frac{1}{2} I \omega^{2}

for a solid cylinder:  I =  \frac{1}{2} m r^{2}
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I will look at the case of a hollow cylinder:

E = mgh + I \omega ^{2} = constant \\ \\ I =  \frac{mgh}{  \omega^{2} }

That is as far as i get.


7 0
3 years ago
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