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AlekseyPX
3 years ago
7

The world’s largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds. What is the velocity fo

r one of the blades? (THIS IS physics, CIRCULAR MOTION)?
Physics
1 answer:
klemol [59]3 years ago
3 0

Answer:

The free end of the blade has a tangential velocity of about 88.19 m/s

Explanation:

The angular velocity of the blades is  2 \pi /5.7\,\,rad/sec

since the blades are 80 m long, then the tangential velocity of the free end of the blade is:

v_{tan} \approx 88.19\,\,m/s

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Suppose that a certain battery produces a voltage of 1.55V without a load connected (open circuit) and a current of 500mA when s
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Explanation:

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A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If
Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

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E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

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