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3 years ago

Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

Physics

1 answer:

m_a_m_a [10]3 years ago

6 0

Answer:

The peak voltage is 473.86 V

Explanation:

Given;

number of turns of generator, N = 200 turn

diameter of the coil, d = 0.1 m

radius of the coil, r = 0.05 m

Magnitude of the magnetic field, B = 0.8 T

angular frequency, ω = 3600 rpm

angular frequency, ω (rad/s) = $\frac{2\pi}{60} *3600\ rpm = 377.04 \ rad/s$

The peak voltage is given by;

E = NBAω

Where;

A is the area of the coil = πr² = π (0.05)² = 7.855 x 10⁻³ m²

E = (200)(0.8)(7.855 x 10⁻³)(377.04)

E = 473.86 V

Therefore, the peak voltage is 473.86 V

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A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600

3241004551 [841]

Answer:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

$v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}$

Where:

$v_{o}$ - Initial speed of the rocket, in m/s.

$v_{ex}$ - Exhaust gas speed, in m/s.

$m_{o}$ - Initial total mass of the rocket, in kg.

$m$ - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

$m(t) = m_{o} + r\cdot t$

The initial total mass of the rocket is:

$m_{o} = 750\,kg$

The fuel consumption rate is:

$r = -\frac{600\,kg}{30\,s}$

$r = -20\,\frac{kg}{s}$

The function for the current total mass of the rocket is:

$m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t$

The speed function of the rocket is:

$v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}$

The speed of the rocket at given instants are:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

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A batter hits a baseball so that it leaves the bat with an initial speed of 60m/s at an initial angle of 42 with the horizontal

Elza [17]

Answer: 24.06°

Explanation:

So,we can say after t if it reaches height h then,

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Given t = 2s

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r = 37 cos 53 *2 = 44.52m

So,after 2s the baseball will be lying 39.57m

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So, Uy = 37 sin 53- 9.8* 2

or, U = 9.95m/s

And.horizontal component of velocity remains

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So.magnitude of velocity after 2s is

Square root of (Vx^2 + Vy^2)= 24.4m/s

Making an angle of tan 22.27/9.95 = 24.06°

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3 years ago