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Reika [66]
3 years ago
5

Are rubber bands and metal springs the same? Explain your reasoning​

Physics
2 answers:
leonid [27]3 years ago
7 0

Answer:

No, they are not same.

Explanation:

The main reason for the difference is that <em>you are deforming the rubber band more than the </em><em>spring.</em>

d1i1m1o1n [39]3 years ago
4 0

Answer:

No, they are not because when you stretch the rubber band you are deforming it but with the spring it is made to go up and down.

Explanation:

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When is an atom stable? A. when it has a full outer orbit. B. when it has the same amount of elections as protons. C. when it ha
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An electron is moving the east with a speed of 5.0 × 106 m/s. There is an electric field of
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The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.

<h3>What is Electric field?</h3>

Electric field is the physical field that surrounds a charge.

<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
  • From Newton's second law, the acceleration the electron will be

a=F/m=qE/m

  • where q= charge of electron

E= electric field

m= mass of electron

=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)

=10¹⁵×0.526m/s²

  • The kinematics equation v²=v0²+2a(Δx)
  • where v=final velocity of the electron

v0=initial velocity of the electron =5×10⁶m/s

a=acceleration of the electron =10¹⁵×0.526m/s²

Δx=distance moved by the electron in east direction =1cm=10^-2m

  • Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2

=25×10¹²+10.52×10¹²

=35.52×10¹²

  • Now velocity of electron=5.95×10⁶m/s.

Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.

Learn more about electric field here:

brainly.com/question/26199225

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5 0
2 years ago
A 0.18 m radius pulley is free to rotate about a horizontal axis. A mass and a mass are attached by a massless string, which is
Musya8 [376]

Answer:

T = 1.766(M-m) Nm where M and m are the 2 masses of the objects

Explanation:

Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.

Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m

T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm

8 0
3 years ago
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