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pickupchik [31]
3 years ago
5

What would happen if you took a really strong metal box and you started teleporting in hundreds and hundreds of small cotton bal

ls until there is no space left and when you are 100% sure there is no more space left u teleport in a small metal ball.
What would happen to the metal box?Would it explode?Or what would happen to it?

Ps:the box was created as just one pice so it doesn’t have any lid or anything.And I know teleportation is still not possible.
Physics
2 answers:
mina [271]3 years ago
5 0
It's your story. You invented teleportation to make an exciting plot. Now go ahead and finish the story, by making up the thrilling climax.
mojhsa [17]3 years ago
4 0
It will not explode since the mass of the cotton balls is so low but rather will most likely break the lock and hinges and come out.
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A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn
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Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

        F = q vB

the acceleration of the particle is centripetal

        a = v² / r

we substitute

        qvB = m v² / r

         qBr = m v

          q =\frac{m\  v}{B\  r}

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

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