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3 years ago

Imagine that you are approaching a black hole in a spacecraft. What would you see? What would happen to you?

Physics

2 answers:

aleksandr82 [10.1K]3 years ago

5 0

Answer:

When i look outside the black hole things will look warped.

Explanation:

What I would see depends on the size of the black hole.

If the black hole were small I would die before entering the black hole. I would die even before entering the event horizon. As the tidal forces of a small black are large because of the event horizon being so close to the center of the black hole my body would spaghettifi (being stretched to a small strand like spaghetti) before entering the black hole. The light would bend due to gravitational lensing.

Now, if I were to enter a supermassive black hole then i would be able to enter the event horizon because of its distance from the center of black hole. The light reaching my eyes would bend drastically. The universe outside would look as if it were warped due to gravitational lensing. Then no light would enter my eye as the mass of the black hole consumes me.

HACTEHA [7]3 years ago

3 0

As you approach the event horizon, your body would be pulled towards the black hole. It's a process called sphagettification.

Here's a quote from an article explaining this.
And gravity from the black hole is starting to pull on your feet more than your head. "The gravity wants to sort of stretch you in one direction and squeeze you in another," says Joe Polchinski.<span>
</span>
As for what you see, that is hard to explain without showing a video, at least for me, but you would see this black hole (ha) and whatever is outside starts to flatten and condense (from your POV) then you would fall into the black hole and then it's complete blackness.

But all of this is just educated guessing, and honestly, you'd be dead before anything would happen.

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pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

$F = mg$

now the acceleration is given as

$F = ma = mg$

$a = g$

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

3 0

3 years ago

(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o

IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200

SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

1 and your displacement from the store to your friend’s house

be Dd

Given: Dd

1 = 200 m [N]; Dd

2 = 600 m [S]

Required: Dd

Analysis: Dd

T 5 Dd

1 1 Dd

Solution: Figure 6 shows the given vectors, with the tip of Dd

joined to the tail of Dd

2. The resultant vector Dd

T is drawn in red,

from the tail of Dd

1 to the tip of Dd

2. The direction of Dd

T is [S].

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

T is 400 m.

Statement: Relative to your starting point at your home, your

total displacement is 400 m [S].

6 0

2 years ago

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon’

Yuki888 [10]

Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

$s = ut + \frac{1}{2}at^2$

where u = initial velocity = 0 m/s

a = acceleration = $250 m/s^2$

t = time = 0.02 s

Therefore:

$s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m$

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

$distance = speed * time$

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

$v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s$

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

8 0

3 years ago

A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string

Darya [45]

Answer:

The speed of wave is $v_1 = 129.1 \ m/s$

The new speed of the two waves is $v = 129.1 \ m/s$

Explanation:

From the question we are told that

The mass of the string is $m = 60 \ g = 60 *10^{-3} \ kg$

The length is $l = 2.0 \ m$

The tension is $T = 500 \ N$

Now the velocity of the first wave is mathematically represented as

$v_1 = \sqrt{ \frac{T}{\mu} }$

Where $\mu$ is the linear density which is mathematically represented as

$\mu = \frac{m}{l}$

substituting values

$\mu = \frac{ 60 *10^{-3}}{2.0 }$

$\mu = 0.03\ kg/m$

$v_1 = \sqrt{ \frac{500}{0.03} }$

$v_1 = 129.1 \ m/s$

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

8 0

3 years ago