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3 years ago

Force causes __________ dimensional objects to rotate

1 answer:

3 0

Forces may cause a three-dimensional object to rotate. These forces are called "orthogonal force vectors" which composes of a horizontal component and vertical component, which hand-in-hand, help a body to rotate in accordance with the force applied on a part of its surface.

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A 0.10-kg cart traveling in the positive x direction at 10.0 m/s collides with a 0.30-kg cart at rest. the collision is elastic.

yuradex [85]

By definition for an elastic shock, we have that if the second particle is at rest, then the final velocity of the first particle will be given by vf1 = ((m1-m2) * vi1) / (m1 + m2). Then, substituting the values: vf1 = ((0.1-0.3) * 10) / (0.1 + 0.4) = - 5m / s.

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3 years ago

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What is the name of alfred wegener's hypothesis about moving land masses?

Umnica [9.8K]

Pangaea; <span>the Greek word "pan" means "all" and Gaea or Gaia (or Ge) was the Greek name of the divine personification of the Earth. </span>

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3 years ago

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The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of

Aleks04 [339]

Answer:

(a) 1, average velocity = -65.6 m/s

2, average velocity = -64.8 m/s

3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given by;

$y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}$

(1)

$y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s$

(2)

$y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s$

(3)

$y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s$

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

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4 years ago

Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual

Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será: