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NNADVOKAT [17]
3 years ago
5

Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth? Assume that t

he orbits are circular and both satellites have the same mass. (b) Which satellite orbits with a longer period, T, around earth? Assume that the orbits are circular and both satellites have the same mass. A B
Physics
1 answer:
alexandr1967 [171]3 years ago
3 0

Answer:

a) the one with a lower orbit    b) the one with a higher  orbit

Explanation:

Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.

In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.

\frac{v^2}{r} = \frac{GM}{r^2}.

so the velocity is

v = \sqrt{\frac{GM}{r} }

where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.

The orbital period of any object in orbit is

T = 2\pi \sqrt{\frac{a^3}{GM} }

where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.

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2. A 4.0 kg magnetic toy car traveling at 3.0 m/s east collides and sticks to a 5.0 kg toy magnetic car also traveling at 2.0 m/
Afina-wow [57]

Answer:

2.44 m/s due East

Explanation:

From the question given above, the following data were obtained:

Mass of 1st car (m₁) = 4 Kg

Velocity of 1st car (u₁) = 3 m/s

Mass of 2nd car (m₂) = 5 Kg

Velocity of 2nd car (u₂) = 2 m/s

Final velocity (v) =?

The final velocity can be obtained as follow:

v(m₁ + m₂) = m₁u₁ + m₂u₂

v(4 + 5) = (4×3) + (5×2)

9v = 12 + 10

9v = 22

Divide both side by 9

v = 22/9

v = 2.44 m/s

Thus, the final velocity is 2.44 m/s.

Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.

4 0
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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