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Nataliya [291]
3 years ago
15

Which is correct about the charge movement when a battery is connected to copper wire?a.)Electrons leave the positive battery te

rminal and electrons enter the negative terminal. b.)Electrons leave the negative battery terminal and electrons enter the positive terminal. c.)Protons leave the positive battery terminal and positive charges enter the negative terminal. d.)Protons leave the negative battery terminal and positive charges enter the positive terminal.
Physics
1 answer:
professor190 [17]3 years ago
5 0

Answer:

b.)Electrons leave the negative battery terminal and electrons enter the positive terminal.

Explanation:

- First of all, the carriers of charge in an electrical circuit are the electrons, not the protons. In fact, electrons in a conductor (such as the copper wire) are free to move, while protons are bond inside the nuclei of the atoms, so they are not free to move. This means that only choices a) and b) could be correct.

- Finally, we know that charges with same sign repel each other, while charges with opposite sign attract each other. This means that electrons are repelled by the negative terminal of the battery and attracted by the positive terminal of the battery. Therefore, the correct choice must be

b.)Electrons leave the negative battery terminal and electrons enter the positive terminal.

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The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15
Lady bird [3.3K]

Answer:

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

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3 years ago
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Marysya12 [62]

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A 3858 N piano is to be pushed up a(n) 3.49 m frictionless plank that makes an angle of 31.6 ◦ with the horizontal. Calculate th
uranmaximum [27]
The work done will be equal to the potential energy of the piano at the final position

P.E=m.g.h

.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
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sinx=opposite side / hypotenuse
= h/L

then h=L.sinx=3.49×sin31.6°=0.638m

weight w=m.g
m=w/g=3858/10=385.8kg

Consider Gravity g=10m/s2

then P.E.=m.g.h=385.8kg×10×0.638=2461.404J

then Work W=P.E.=2451.404J
8 0
3 years ago
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1. white
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3. reflecting
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What happens to a sheet of copper as kinetic energy of copper molecules decrease
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