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Rzqust [24]
4 years ago
6

1. Dos puntos A y B están separados en línea recta 1475 m. Por el punto A pasa hacia el punto B un móvil con una velocidad de 15

m/s. Cinco segundos más tarde pasa por el punto B hacia el punto A otro móvil con una velocidad de 25 m/s. Determinar
a) El tiempo que se mueve cada uno hasta que se cruzan.
b) El espacio que recorre cada móvil hasta que se cruzan.
Physics
1 answer:
ikadub [295]4 years ago
6 0

Answer:

a) 35 sequndos   b) M1 = 600m   M2 = 875m

Explanation:

a)

Antes de que el otro móvil comience a moverse, el primer móvil ha viajado 5 segundos a una velocidad de 15 m / s.

5 segundos x 15 m / s = 75 metros.

Podemos establecer una ecuación con la información que conocemos. Juntos, en total, los dos recorren 1475 metros.

Voy a representar el tiempo como t. Un móvil ya ha recorrido 75 metros y hará otros 15 t. El otro aún no se ha movido y viajará 25t. Esos tres números se sumarán a 1475.

75 + 15t + 25t = 1475.

Resta 75 de ambos lados y suma términos similares de t: 40t = 1400

t = 35 segundos.

b)

Ahora tome la distancia recorrida por cada vehículo por separado y conecte 35 segundos para t.

Primer móvil: 75 + 15t = 75 + 15 (35) = 600m

Segundo móvil: 25t = 25 (35) = 875m

Usé un traductor para entender la pregunta y luego traduje lo que escribí. Lo siento si mi explicación no es clara.

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Answer:

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Explanation:

To find the matrix, let's decompose the vectors, the rotated angle is (-60C) for the prime system

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the transformation system is

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the transformation matrix is

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A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara
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Answer:

(a)\alpha = 53.73 m/s^2

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(c)\tau = 428 \times 53.73  = 22996 .44Nm

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

\omega = \frac{v}{l}

\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s

for  angular acceleration, use the following equation.

\omega _{f}^2 = \omega_{i}^2+2\alpha\theta

since \omega _{i} = 0

here  for one circle is 2 π radians.   therefore for one quarter of a circle is π/2 radians

so   for one quarter \theta = \frac{\pi }{2}

(12.99)^2 = 2\alpha(\frac{\pi }{2})

on solving

\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2

(b)

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I = MR^2

I = 18.2 \times 14.51

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so total moment of inertia =  428 kgm^2

(c)

\tau = I\alpha

\tau = 428 \times 53.73  = 22996 .44Nm

3 0
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