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Ainat [17]
1 year ago
7

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

great amount of angular momentum when rotating. A wind turbine has a total of 3 blades. Each blade has a mass of m = 5500 kg distributed uniformly along its length and extends a distance r = 44 m from the center of rotation. The turbine rotates with a frequency of f = 13 rpm.
a. Calculate the total moment of inertia of the wind turbine about its axis, in units of kilogram meters squared.
b. Calculate the angular momentum of the wind turbine, in units of kilogram meters squared per second.
Physics
1 answer:
horsena [70]1 year ago
7 0

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵  kgm² s⁻¹  

Explanation:

a)Given

m = 5500 kg

l = 44 m

Moment of inertia of one blade

I= 1/3 x m l²

where m is mass of the blade

l is length of each blade.

Putting all the required values, moment of inertia of one blade will be

I= 1/3 x 5500 x 44²  

I= 35.49 x 10⁵ kg.m²

Moment of inertia of 3 blades

I= 3 x 35.49 x 10⁵ kg.m²

I= 106.48 x 10⁵ kg.m²

b) Angular momentum 'L' is given by

L =I x ω

where,

I= moment of inertia of turbine i.e  106.48 x 10⁵ kg.m²

ω=angular velocity =2π f

f is frequency of rotation of blade i.e  13 rpm

f = 13 rpm=>= 13 / 60 revolution per second

ω = 2π f =>  2π  x  13 / 60 rad / s

L=I x ω =>106.48 x 10⁵ x   2π  x  13 / 60

  = 144.97 x 10⁵  kgm² s⁻¹    

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avanturin [10]

Answer:

\dfrac{dh}{dt} =5\ ft/s

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300  tanθ

\dfrac{dh}{d\theta } = 300 sec^2\theta

now  \dfrac{dh}{dt} can be written as

\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}

\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}

When θ = π/4,

\dfrac{dh}{d\theta } = 300 sec^2\theta

\dfrac{dh}{d\theta } = 600

\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}

\dfrac{dh}{dt} =600\times \dfrac{1}{120}

\dfrac{dh}{dt} =5\ ft/s

5 0
2 years ago
How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236
Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between  non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

 

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

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4 0
2 years ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

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m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

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W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

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∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

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∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

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Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

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v_{o} =\sqrt{14.472}

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2 years ago
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makkiz [27]
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t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
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t1 = (56+24)/32 = 2.5 
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3 0
2 years ago
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Katyanochek1 [597]

Answer:

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5 0
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