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Elenna [48]
2 years ago
13

What happens to the Total Energy as the spring bounce?

Physics
1 answer:
fiasKO [112]2 years ago
4 0
Total energy will remain the same
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A current of 0.4 A flows through a wire. How many electrons flow through a cross section of
Free_Kalibri [48]

9 × 10²¹ electrons flow through a cross section of the wire in one hour.

<h3>What is the relation between current and charge?</h3>
  • Mathematically, current = charge / time
  • In S.I. unit, Charge is written in Coulomb and time in second.

<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
  • Charge= current × time
  • Current= 0.4 A, time = 1 hour= 3600 s
  • Charge= 0.4× 3600

= 1440 C

<h3>How many numbers of electrons present in 1440C of charge?</h3>
  • One electron= 1.6 × 10^(-19) C
  • So, 1440 C = 1440/1.6 × 10^(-19)

= 9 × 10²¹ electrons

Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.

Learn more about current here:

brainly.com/question/25922783

#SPJ1

4 0
2 years ago
In a machine, work output is less than work input because some energy is converted into thermal energy. true or false.
tamaranim1 [39]
True ..........................
7 0
3 years ago
Read 2 more answers
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
2 years ago
Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp
Thepotemich [5.8K]
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!

<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
3 0
3 years ago
One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e
frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is 1.122 X 10^{-7}  kg

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

E= \Delta mc^{2}

where \Delta m = change in mass

c = speed of light = 3 \times 10 ^{8}m/s

Making m the subject of the formula, we can find the change in mass to be

\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0
3 years ago
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