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MArishka [77]
3 years ago
12

If the momentum of a system is to be conserved, which must be true of the net external force acting on the system?

Physics
1 answer:
kari74 [83]3 years ago
7 0

Answer:

D. zero

Explanation:

For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.

There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.

F_{ext} = 0

Therefore, the correct option is "D"

D. zero

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Explanation:

1. - right to vote

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3 years ago
4.- Una vagoneta de 1000 kg de peso parte del reposo en el punto 1 y desciende, sin rozamiento, por la vía indicada en la figura
Akimi4 [234]

Answer:

A) 49,050 N

B) 16 m

Explanation:

Question:

El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.

La masa dada del vagón, m = 1,000 kg

La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m

A) El radio en el punto 2, el punto más bajo, R = 6 m

La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;

N = El peso del vagón + La fuerza de movimiento del vagón

∴ N = m × g + m × a

Dónde;

g = La aceleración debida a la gravedad ≈ 9,81 m / s²

a = La aceleración del vagón

Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R

∴ m × a = m × v² / R

Dónde;

v² = La velocidad del vagón en el punto 2 = 2 · g · h₁

Por lo tanto;

N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R

∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050

La fuerza que ejerce el vagón en el punto 2, N = 49,050 N

B) En el punto 3, tenemos;

N = m · g - m · a₃

La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃

∴ La altura en el punto 3, h₃ = 4 m

El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)

Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando

Por lo tanto;

N = m · g - m · a₃ = 0

m · g = m · a₃ = m · v₃² / R₃ = m · (2 ​​· g · (h₁ - h₃)) / R₃

∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16

El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.

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antoniya [11.8K]

Answer:

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