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wariber [46]
4 years ago
10

When current is flowing in an ordinary metal wire, which is closest to the average drift speed of the electrons?

Physics
1 answer:
Luba_88 [7]4 years ago
7 0

Answer:

Option C is correct.

1mm/s

Explanation:

Drift velocity for electrons in wires is the average velocity attained by charged electrons in a material due to an electric field or their average velocity of the electrons during the conduction of electricity.

It is given mathematically as

v = I/(nAq)

where v = drift velocity

I = current in the wire

n = number of electrons per cubic metre for the wire

A = Cross sectional Area of the wire

q = charge on one electron

Electrons normally move with speed close to the speed of light in vacuums but the drift velocity of an electron in conductors is usually very small (mm/s range!).

This is because when electrons move randomly towards higher potential, electrons continuously collide and scatter off of crystal defects, phonons, impurities and vacancies in the conductor in a random fashion. Due to such collisions, electron lose some of their kinetic energy. As a result, electrons do not accelerate but travels with a finite average velocity which we refer to as a drift velocity.

Although as soon as the electric field is established in the conductor, the current starts flowing inside the conductor at the speed of light and not at the speed at which the electrons are drifting.

For example, we will pick common, relatable values for these parameters.

Drift velocity is given mathematically as

v = I/(nAq)

where v = drift velocity

I = current in the wire = assume 10 A

n = number of electrons per cubic metre for the wire = assume the conductor is copper, n = 8.5 × 10²⁸ electrons/m³

A = Cross sectional Area of the wire = assume wire of radius 1mm, A = πr² = 3.142 × 10⁻⁶ m²

q = charge on one electron = 1.602 × 10⁻¹⁹ C

v = (10)/(8.5 × 10²⁸ × 3.142 × 10⁻⁶ × 1.602 × 10⁻¹⁹)

v = 2.34 × 10⁻⁴ m/s = 0.234 mm/s which shows the value is closest to 1 mm/s

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sashaice [31]

Answer:

The answer is: Increase the Number of Ropes/Pulleys.

Explanation:

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This machine is used <em>in order to decrease the amount of time it takes to do work, such as lifting of objects.</em>

Remember that "mechanical advantage"<u> refers to the ratio between the force applied to a machine and the force produced by it.</u> So, when it comes to pulleys, the mechanical advantage increases when the number of ropes are also increased. They are directly proportional. This goes the same way with increasing the number of pulleys. Once you increase the number of pulleys, the mechanical advantage also increases.

8 0
3 years ago
1 pts
Nuetrik [128]

Answer:

The leaf cells would be unable to convert the sun's energy to make food.

Explanation:

The chloroplast in the leaf are used to make food for the plant. If the leaf loses its chloroplast, the leaf with not be able to use solar energy to manufacture their food.

  • During photosynthesis, green plants manufacture their food.
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4 0
3 years ago
A 5 kilogram cat is resting on top of a bookshelf that is 3 meters high. What is the cat’s gravitational potential energy relati
Free_Kalibri [48]
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5 0
3 years ago
A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the
Naily [24]

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

r = \sqrt{(3.5)^2+(2.5)^2}

  =\sqrt{8.5}

SL=85

As we know,

⇒  SL=10 \ log(\frac{I}{I_o} )

     85=10 \ log(\frac{I}{10^{-12}} )

      I=3.162\times 1^{-4} \ W/m^2

Now,

⇒  P_{front} = I(2\pi r^2)

               =(3.162\times 10^{-4})(2\pi\times 18.5)

               =0.0368 \ W

               =0.75 \ P

or,

                =0.049 \ W

7 0
3 years ago
The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord
SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

P_x=rcos\theta

P_y=rsin\theta

where \thetais angle made by horizontal bar with x axis

Velocity at y=150 mm

150=300sin\theta

thus \theta =30^{\circ}

position ofP_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}

P_x=259.80 mm

P=259.80\hat{i}+150\hat{j}

Velocity at this instant

u_x=-rsin\theta =300\times sin30=-150 mm/s

u_y=rcos\theta =300\times cos30=259.80 mm/s

4 0
3 years ago
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