How did Orion become a constellation?

1 answer:

5 0

While the Orion constellation is named after the hunter in Greek mythology, it is anything but stealthy. Orion, which is located on the celestial equator , is one of the most prominent and recognizable constellations in the sky and can be seen throughout the world.

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An object with a mass of 5.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the acceleration? (20 poin

Readme [11.4K]

Answer:

40N

Explanation:

F=ma

F= 5 × 8 = 40N

.......

4 0

3 years ago

For the simple harmonic motion equation

My name is Ann [436]

Given that : d = 5sin(pi t/4), So, maximum displacement, d = 5*(+1) = 5 Also, maximum displacement, d = 5*(-1) = -5

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4 years ago

Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the f

Alex17521 [72]

Answer:

g' = g/4

Explanation:

The value of the free-fall acceleration at the surface of the earth, can be obtained applying Newton's 2nd law, assuming that the only force acting on an object at the surface of the earth, is the one produced by the mass of the Earth, i.e. gravity.
This force can be expressed according the Newton's Universal Law of Gravitation , as follows:

$F_{g} = G*\frac{m_{x} *m_{E} }{r_{E}^{2} } (1)$

From Newton's 2nd Law, we have:
$F = m* a (2)$
Since the left sides in (1) and (2) are equal each other, both right sides must be equal each other also.
Simplifying the mass m, we can write the acceleration a in (2) as the acceleration due to gravity, g, as follows:

$g = G*\frac{m_{E} }{r_{E}^{2} } (3)$

Since G is an universal constant, and the mass mE remains constant, if we double the radius of the Earth, the new value for the acceleration due to gravity (let's call it g'), is as follows:

$g' = G*\frac{m_{E} }{(2r_{E})^{2} } =G*\frac{m_{E} }{4*r_{E}^{2} } = g*\frac{1}{4} =\frac{g}{4} (4)$