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2 years ago

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A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

of 837 rpm. The radius of the turbine generator is 0.62 m and its rotational acceleration is 5.9 rad/s2 . What is the turbine’s angular displacement (in radians) after 3.2 s?

1 answer:

Salsk061 [2.6K]2 years ago

3 0

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

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Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

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There is a rate of change of the perpendicular component of momentum.

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We need to calculate the magnitude of the rate of change of the child's momentum

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Put the value into the formula

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Electric charge on the plastic cube: $1.3\cdot 10^{-7}C$

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

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Since the voltage is inversely proportional to r, we can use:

$Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V$

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

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a bouncing ball --> YES, because the ball is moving AND it is at a certain height above the ground

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