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LuckyWell [14K]
3 years ago
8

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

617 rev/s. A point on one of the blades traces out acircle with a radius of 0.020m as the blade rotates. a) What is themagnitude of the centripetal acceleration that the blade mustsustain at this point? b) Express this acceleration as amultiple of g = 9.80m/s2 .
answer: 3.0*105m/s2
Physics
1 answer:
kvasek [131]3 years ago
8 0

Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

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θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

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α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

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θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

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t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

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22π=21.26t

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t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
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h=.5Gt2 

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Answer:

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