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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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If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 50.5 m/s in the opposite directi
Naya [18.7K]

Answer:

The magnitude of change in momentum of the ball is 97.5 m and impulse is also 97.5 m

Explanation:

Given:

Velocity of a pitched ball v _{i} = 47 \frac{m}{s}

Velocity of ball after impact v_{f}  = -50.5 \frac{m}{s}

From the formula of change in momentum,

  \Delta P = m (v_{f} -v_{i}  )

Here mass is not given in question,

Mass of ball is m

Change in momentum is given by,

\Delta P = m (-50.5 -47)

\Delta P = -97.5 m

Magnitude of change in momentum is

\Delta P = 97.5 m

And impulse is given by

 J = \Delta P

J = -97.5 m

So impulse and

Therefore, the magnitude of change in momentum of the ball is 97.5 m and impulse is also -97.5 m

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3 years ago
What makes the ball stop on rolling after somtime​
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Describe how to use a reference point and a set of reference directions to define the position of an object.
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You obtain a spectrum of an object in space. The spectrum consists of a number of sharp, bright emission lines. Is this object a
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4. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm to 400 nm. It is not
vampirchik [111]

The ranges of frequency are:

UVA: [7.50-9.38]\cdot 10^{14} Hz

UVB: [9.38-10.71]\cdot 10^{14} Hz

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is the following:

f = \frac{c}{\lambda}

where

f is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the UVA, the range of wavelength is 320 nm - 400 nm, so

\lambda_1 = 320 \cdot 10^{-9} m\\\lambda_2 = 400 \cdot 10^{-9} m

So the corresponding frequencies are

f_1 = \frac{3\cdot 10^8}{320\cdot 10^{-9}}=9.38\cdot 10^{14} Hz\\f_2= \frac{3\cdot 10^8}{400\cdot 10^{-9}}=7.50\cdot 10^{14} Hz

For the UVB, the range of wavelength is 280 nm - 320 nm, so

\lambda_1 = 280 \cdot 10^{-9} m\\\lambda_2 = 320 \cdot 10^{-9} m

So the corresponding frequencies are

f_1 = \frac{3\cdot 10^8}{280\cdot 10^{-9}}=1.07\cdot 10^{15} Hz\\f_2= \frac{3\cdot 10^8}{320\cdot 10^{-9}}=9.38\cdot 10^{14} Hz

So the ranges of frequency are:

UVA: [7.50-9.38]\cdot 10^{14} Hz

UVB: [9.38-10.71]\cdot 10^{14} Hz

Learn more about waves, frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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