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postnew [5]
3 years ago
7

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

. A) Calcular la fuerza que la vía ejerce sobre la vagoneta en el punto 2, donde el radio de curvatura es de 6 m. B) Determinar el mínimo valor del radio de curvatura en el punto 3 para salvar dicho punto
Physics
1 answer:
Akimi4 [234]3 years ago
5 0

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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3 years ago
Select all that apply to electrons and energy levels. - Scientists have not yet determined exactly why electrons do not collapse
Elenna [48]

Answer:

- The limitation of the maximum number of electrons in a given energy level can be used to account for the periodic recurrence of properties as the number of electrons increases.

Explanation:

First - Scientists have not yet determined exactly why electrons do not collapse into the nucleus. FALSE: Scientists do know why electrons do not collapse. Since the beginins of quantum mechanics it's known that the energy at small scales is quantized, that means there only can be certain values meaning that the energy do not change continously. In the case of the electron, it can only have certain levels of energy, that means they do not radiate continously as the go arround the atom, instead it is only allowed to have a certain amount of energy in a given state therefore it can not lose energy continously collapsing into the nucleus.

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7 0
3 years ago
Jessica is running a 10K. She alternates between running and walking each kilometer. She runs at a rate of 1 kilometer every 5 m
saveliy_v [14]

Answer:

c.100 minutes

Explanation:

Total distance = 10 km

Runs for 1 km every 5 minutes

walks 1 km every 15 min

She alternates between walking and running  so,  Jessica will walk 5 km and run 5 Km

Time taken by Jessica for walking : 5 km

Time taken to walk 1 km=5 minutes

Time taken to walk 5 km

=> 5 X 5

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=> 5 X 15

=>75 minutes

Total time taken   = Time taken by Jessica for walking + Time taken by Jessica for Running

=>25 minutes +75 minutes

=> 100 minutes

8 0
3 years ago
Read 2 more answers
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