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Gnoma [55]
2 years ago
11

Problema 1. Un Clavadista se lanza desde un trampolín a diferentes alturas 10 m, 3 m, y 1 m, Calcular:

Physics
1 answer:
Salsk061 [2.6K]2 years ago
4 0

La velocidad del buzo es de 7.04 m/s, 8.84 m/s y 2.27 m/s respectivamente.

<h3 /><h3>Velocidad </h3>

La velocidad es la relación entre la distancia total recorrida y el tiempo total empleado. Está dado por:

Velocidad = distancia/tiempo

Para 10m de altura:

  • Velocidad = 10 m/1.42 s = 7.04 m/s

Para 3m de altura:

  • Velocidad = 3 m/0.78 s = 3.84 m/s

Para 1m de altura:

  • Velocidad = 1 m/0.44 s = 2.27 m/s

La velocidad del buzo es de 7.04 m/s, 8.84 m/s y 2.27 m/s respectivamente.

Obtenga más información sobre la velocidad en: brainly.com/question/4931057

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What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the
timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B =  $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $

Therefore, the maximum initial K.E. = 3066.67 J

3 0
2 years ago
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