PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

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

viva [34]

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

6 0

1 year ago

Where are the life forms of the biosphere located

bulgar [2K]

<span>The life forms of the biosphere are located within Earth's surface.</span>

7 0

2 years ago

While playing Quidditch you throw the quaffle straight down to the person below you at

oee [108]

Answer:

the ans will be because it has 1.672

3 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$