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OLEGan [10]
3 years ago
7

An engineer is testing a large wind turbine that is used generating energy. Intially the large wind blades are locked in place a

nd not rotating. In a steady wind, the engineer releases the blades and they begin to rotate. She notes from the instruments that for the first 60 seconds the blades have a constant angular acceleration of 0.05rad/s2 . How many complete revolutions of the turbine occur during this time
Physics
2 answers:
soldier1979 [14.2K]3 years ago
5 0

Answer:

28 revolutions

Explanation:

Here we are going to use the accelerated angular motion formulas.

θ=wo*t+\frac{1}{2}*\alpha*t^2

because it starts from rest, wo=0

θ=\frac{1}{2}*(0.05rad/s^2)*(60s)^2

θ=180rad

The number of revolutions is given by:

N=\frac{180rad}{2*\pi\frac{rad}{rev}}=28.64rev

So the turbine completed 28 revolutions at that time.

emmasim [6.3K]3 years ago
4 0

Answer:

3 revolution in rad

Explanation:

angular acceleration α= rev/time

but α= 0.05rad/s

rev = α * time(in sec)

rev = 0.05 * 60

rev = 3rad/s

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A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
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Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

5 0
3 years ago
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