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kvv77 [185]
4 years ago
12

Think about this hypothetical (not real) situation: Astronaut Dustin and Astronaut Barb are floating in space. The force of grav

ity between Astronaut Dustin and Planet Demi
120,265 Newtons. The force of gravity between Astronaut Barb and Planet Demi is 354,999 Newtons. What can you infer about the relative DISTANCE of each astronaut
from planet Demi? Explain your answer. ​
Physics
1 answer:
Maslowich4 years ago
6 0

The distance between Dustin and the planet is larger than the distance between Barb and the planet

Explanation:

The magnitude of the gravitational force between each astronaut and the planet is given by

F=G\frac{Mm}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

M is the mass of the planet

m is the mass of the astronaut

r is the separation between the astronaut and the planet

In this problem, we have:

  • The force of gravity between Dustin and the planet is 120,265 N
  • The force of gravity between Barb and the planet is 354,999 N

We see that the force exerted by the Planet on Barb is much greater than the force exerted by the planet on Dustin. Assuming that the mass of Dustin and Barb is similar, then we can say that the magnitude of the force of gravity depends mainly on the distance:

F=\frac{1}{r^2}

And since the force is inversely proportional to the square of the distance, this means that the distance between Dustin and the planet is larger than the distance between Barb and the planet.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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r = n² · α₀

where:
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Since d = 2 · r, we can write:
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The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen
Eddi Din [679]

Answer:

\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }

Explanation:

Additional information:

<em>The ball has charge </em>-q_0<em>, and the ring has  positive charge </em>+Q<em> distributed uniformly along its circumference. </em>

The electric field at distance z along the z-axis due to the charged ring is

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Therefore, the force on the ball with charge -q_0 is

F=-q_oE_z

F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}

and according to Newton's second law

F=ma=m\dfrac{d^2z}{dz^2}

substituting F we get:

- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}

rearranging we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0

Now we use the approximation that

z^2+a^2\approx a^2 <em>(we use this approximation instead of the original </em>d^2+a^2\approx a^2<em> since </em>z<em>, our assumption still holds )</em>

and get

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0

Now the last equation looks like a Simple Harmonic Equation

m\dfrac{d^2z}{dz^2}+kz=0

where

\omega=\sqrt{ \dfrac{k}{m} }

is the frequency of oscillation. Applying this to our equation we get:

m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}

\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}

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