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julia-pushkina [17]
3 years ago
8

What is the law of conservation of energy.

Physics
1 answer:
Sever21 [200]3 years ago
3 0

The law of conservation of energy implies that energy can neither be created nor destroyed, but can be changed from one form to another.

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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
Why is it important for scientist to do research?
sweet-ann [11.9K]

Answer:

Because if they dont research first they will be unprepared

Explanation:

3 0
3 years ago
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How do boron-10 and boron-11 differ?
tensa zangetsu [6.8K]

Answer:

I think it has to do something with their ionizations... not entirely sure though.

Explanation:

3 0
3 years ago
The difference in a brass rod and steel rod is same as 12cm at all temperature .Find the length of the rod at 0°C.​
egoroff_w [7]

Answer:

I think its 12 cm itself. Because in the question it is said that the length is same in all temperature

5 0
3 years ago
Ml(d^2θ/dt^2) =-mgθ
Nata [24]

The equation of motion of a pendulum is:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,

where \ell it its length and g is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (\theta \ll 1), we can use:

\sin\theta \simeq \theta.

Additionally, let us define:

\omega^2\equiv\dfrac{g}{\ell}.

We can now write:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.

The solution to this differential equation is:

\theta(t) = A\sin(\omega t + \phi),

where A and \phi are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.

This justifies that the period depends only on the pendulum's length.

4 0
3 years ago
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