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horrorfan [7]
2 years ago
15

Please answer the following questions correctly.

Mathematics
1 answer:
mamaluj [8]2 years ago
5 0

Note that a translation along the horizontal is affected by the dependent variable. Hence the correct graphical representation will be Graph C

<h3>Translation and transformation</h3>

Translation is the transformation technique of changing the position of an object on the xy-plane.

Give the parent function f(x) = x^2, if the parent function is translated to the right by 4 units (along the horizontal), the resulting function h(x) will be (x+4)^2

Note that a translation along the horizontal is affected by the dependent variable. Hence the correct graphical representation will be Graph C

Learn more on translation here: brainly.com/question/12861087

#SPJ1

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Gre4nikov [31]
Let x be the volume (liters) of pure (at 100%) acid needed
Let y be the volume (liters) of the other acid (at 10%) needed 
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a) x+ y = 63 liters, AND their respective concentration in acid: is
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3 years ago
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Solnce55 [7]

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;

F =G\times \dfrac{M \cdot m}{r^{2}}

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;

\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2}  = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx  1.14

∴ 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.

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