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slega [8]
3 years ago
5

Is Bytes a scalar or vector?​

Physics
2 answers:
melisa1 [442]3 years ago
8 0
It is a scaler because it’s only fully describes by a magnitude and a numerical alone
Anna007 [38]3 years ago
4 0

Answer:

vector

Explanation:

Is Bytes a scalar or vector?

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Protons, neutrons, electrons, and a nucleus are
tatuchka [14]
It would be Atoms, they’re all made up of these tiny particles
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3 years ago
GIVING BRAINLIEST FIVE STARS AND HEART!
alina1380 [7]

Answer:

A bicycle on the top of the hill has the highest potential energy, and when the bike goes down, it transfers to kinetic because it is moving

Explanation:

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3 years ago
Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.
MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>
  • It is given that, the infinite well having a width of 0.050 mm.
  • We have the expression for energy of electron in an infinite well as,

                  E_n=\frac{n^2h^2}{8mL^2}

  • where;

                m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1

  • Thus, the lowest energy of electron in an infinite well is,

                E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

#SPJ4

7 0
1 year ago
A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic
ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

I_i = Initial angular momentum = 275\ kgm^2

\omega_i = Initial angular velocity = 14 rev/min

I_f = Final angular momentum

\omega_f = Final angular velocity

Here the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s

The final angular velocity is 0.2932 rad/s

7 0
3 years ago
What activity is mainly a strength training excerise?
Sever21 [200]

Biceps curls & pushups , benchpress

6 0
3 years ago
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