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AlekseyPX
3 years ago
15

A high-speed bullet train accelerates and decelerates at the rate of 10 ft/s2. its maximum cruising speed is 120 mi/h. (round yo

ur answers to three decimal places.) (a) what is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
Physics
1 answer:
valina [46]3 years ago
3 0
What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?

d = vo*t + 0.5*a*(t**2)  ..........................(vo= initial speed = 0; a= acceleration; t=time, d = distance)

Vf  = vo  + a*t      (final speed, in this case it will be the cruising speed)


Since vo = 0 (starts from rest)

t = Vf/a

d = 0 + 0.5*a*((Vf/a)**2)

d =  (0.5/a)*(Vf)**2

a = 24545 miles/h2 = 10ft/s2


d1 =  (0.5/24545 miles/h2)*(120 mi/h)**2 = 0.293 miles

From cruising speed, distance traveled after 15 minutes (0.25h)

v = d/t ...................> d2 =v*t = <span>120 mi/h (0.25h)</span> = 30 miles

Maximum distance the train can travel,

D = d1 + d2 = 0.293 + 30  miles = 30.293 miles







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m_w g = (m+m_w)a
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2 years ago
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
ANTONII [103]

Answer:

Explanation:

a )

Each blade is in the form of rod with axis near one end of the rod

Moment of inertia of one blade

= 1/3 x m l²

where m is mass of the blade

l is length of each blade.

Total moment of moment of 3 blades

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ml²

2 )

Given

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c )

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= I x ω

I is moment of inertia of turbine

ω is angular velocity

ω = 2π f

f is frequency of rotation of blade

d )

I = 111 .375 x 10⁵ kg.m² ( Calculated )

f = 11 rpm ( revolution per minute )

= 11 / 60 revolution per second

ω = 2π f

=  2π  x  11 / 60 rad / s

Angular momentum

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Answer:

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