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nata0808 [166]
3 years ago
13

Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of R = 1.22 m and a mass of 67.0 kg . To

prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3510 m/s2 .What is the maximum kinetic energy that can be stored in the fly-wheel?
Physics
1 answer:
eduard3 years ago
5 0

Answer: 71.7 KJ

Explanation:

The rotational kinetic energy of a rotating body can be written as follows:

Krot = ½ I ω2

Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:

Fc = m. ac

It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:

ac =  ω2 r

We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.

As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.

Replacing in the expression for the Krot, we have:

Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ

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Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
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Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

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1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

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6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

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A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh
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Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

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F = 0.24 N

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