Evalúa 13-0{,}75w+8x13−0,75w+8x13, minus, 0, comma, 75, w, plus, 8, x cuando w=12w=12w, equals, 12 y x=\dfrac12x= 2 1 x, equal
1 answer:
Answer:
Step-by-step explanation:
Evaluate 13-0.75w+8x13−0.75w+8x13, minus, 0, point, 75, w, plus, 8, x when w=12w=12w, equals, 12 and x=\dfrac12x= 2 1 x, equals, start fraction, 1, divided by, 2, end fraction.
For this case we have the following expression:
Evaluate for the following values:
w = 12
x = 1/2
We Substituting the values we have:
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Answer:
Only Cory is correct
Step-by-step explanation:
The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;
Where;
G = The universal gravitational constant
M = The mass of one object
m = The mass of the other object
r = The distance between the centers of the two objects
For the gravitational pull of the Earth on a person, when the person is standing on the Earth's surface, r = R = The radius of the Earth ≈ 6,371 km
Therefore, for an astronaut in the international Space Station, r = 6,800 km
The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;
∴ F₁ ≈ 1.14 × F₂
F₂ ≈ 0.8778 × F₁
Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth
However, the International Space Station is moving in its orbit around the Earth at an orbiting speed enough to prevent the Space Station from falling to the Earth such that the Space Station falls around the Earth because of the curved shape of the gravitational attraction, such that the astronaut are constantly falling (similar to falling from height) and appear not to experience gravity
Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.