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pychu [463]
2 years ago
6

You are feeling like spaghetti. Although normally only about 2 meters tall, you are now about 25 meters long. (How fortunate, if

painful, that the being has arranged for your body to become elastic enough so that it is not ripped apart under these conditions.) As you look up over your head, you see things moving pretty quickly in the universe but that lasts only a brief instant, and then all contact with the universe is lost. Where are you
Physics
1 answer:
viva [34]2 years ago
6 0

You are crossing the event horizon of a black hole

When you are feeling like spaghetti and you are normally only about 2 meters tall, you are now about 25 meters long, then look up over your head, you see things moving pretty quickly in the universe but that lasts only a brief instant, and then all contact with the universe is lost, you are crossing the event horizon of a black hole.

<h3>What happens when you are crossing the event horizon of a black hole?</h3>
  • The point of no return is the black hole's event horizon.
  • Anything that continues beyond this point will be absorbed by the black hole and disappear from the known universe forever.
  • The black hole's gravity is so strong at the event horizon that it cannot be overcome or resisted by any mechanical force.

<h3>Is it possible to endure inside an event horizon?</h3>
  • As a result, the individual would survive and gently float over the event horizon of the black hole without being harmed or stretched into a long, thin noodle.

<h3>What occurs beyond the horizon of the event?</h3>
  • A singularity is a truly tiny point that lies beyond the event horizon where gravity is so strong that space-time itself is infinitely bent.
  • The principles of physics as they exist presently break down at this point, making any hypotheses about what lies beyond mere conjecture.

To learn more about black hole visit:

brainly.com/question/27723143

#SPJ4

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lorasvet [3.4K]

Answer:

V_3\approx 4.28\,\,V

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Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega

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and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V

So now we know that the potential drop across the parellel resistors must be:

10 V -  4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps

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Answer:

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