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Colt1911 [192]
3 years ago
5

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

your current weight. Show all work leading to your answer OR describe your solution using 3 -4 complete sentences.
Physics
1 answer:
FrozenT [24]3 years ago
6 0
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
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Explanation:

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Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i
Naddik [55]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

7 0
3 years ago
A lamp is 10% efficient.How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy
k0ka [10]

If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.

20 J/s = 10% of Supply

20 J/s = (0.1) x (Supply)

Divide each side by 0.1:

Supply = (20 J/s) / (0.1)

<em>Supply = 200 J/s  </em>(200 watts)

========================

Here's something to think about:  What could you do to make the lamp more efficient ?  Answer:  Use it for a heater !

If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about.  Suddenly ... bada-boom ... the lamp is 90% efficient !

6 0
3 years ago
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