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jok3333 [9.3K]
3 years ago
10

What type of ions are electrons colliding with to cause resistance in a circuit?

Physics
2 answers:
puteri [66]3 years ago
8 0

Answer:

B. Metallic

Explanation:

It is the metallic ions that offer resistance to the electron flow.

Resistance in a circuit is due to the opposition to the flow of current (flow of electrons) . Resistance depends on the characteristics of the material, like resistivity , length and area of cross section of the metal.

If a material has more resistivity, resistance is more and vice versa. If the length is more or if the length is increased, keeping the area of cross section same, for a given material, the metallic ions colliding with the electrons have to travel a greater distance, there by increasing the resistance.

kkurt [141]3 years ago
6 0
The type of ions that electrons collide with to cause resistance in a circuit is B. Metallic
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A blender takes in 180 J of electrical energy. Of the 180 J, 90 J are transformed into kinetic energy, 60 J are transformed into
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Fa sho it's 30j edge said so
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An airplane flies with a velocity of 55.0 m/s [ 35° N of W] with respect to the air (this is
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Answer:

21 m/s.  

Explanation:

The computation of the wind velocity is shown below:

But before that, we need to find out the angles between the vectors

53° - 35° = 18°

Now we have to sqaure it i.e given below

v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°

v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951

v^2 = 440.6

v = √440.6

v = 20.99

≈ 21 m/s

Hence, The wind velocity is 21 m/s.  

6 0
3 years ago
A vertical wire carries a current straight up in a region where the magnetic field vector points due north. What is the directio
Elanso [62]

Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

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3 years ago
What connects the upper motor neurons to lower motor neurons?
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Answer:

The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

Explanation:

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Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

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3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
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