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kotegsom [21]
2 years ago
10

In order for electrical current to flow through a solid, _____. the solid must have electrons the electrons in the solid must be

able to move the solid must have protons the electrons must be in tight formation
Physics
2 answers:
Luden [163]2 years ago
7 0

<em>The electrons in the solid must be able to move</em>

nydimaria [60]2 years ago
6 0

Electrical current will flow through a solid by the means of electrons.  The electrons in the solid must be able to move from lower to higher electrical potential. The electric current in certain solid electrolytes like ice is entirely composed of flowing ions.

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A resistor resists the flow of electricity and usually converts electrical energy to heat energy. True or False?
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Your answer is: True!
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How can i calculate distances between objects by using the concepts of echo location
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Send wave from your location to the object and wait until echo is back.
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If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return
5 0
2 years ago
Which of the below is an example of mimicry that enables prey species avoid predation?
Marta_Voda [28]

b..a harmless organism imitating the look of a harmful organism

Explanation:

A harmless organism imitating the look of a harmful organism is one example of mimicry that enables prey species avoid predation.

Prey are smaller and less harmful organisms often hunted by larger organisms usually carnivores.

Mimicry is a form of evolutionary adaptation process in which two organisms of the same specie or different species tends to look alike.

It is a subtle defense mechanism developed by organism over an extended period of time.

learn more:

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6 0
3 years ago
In a summer storm, the wind is blowing at a velocity of 8 m/s north. Suddenly in 3 seconds, the winds velocity is 23 m/s north.
cricket20 [7]

Answer:a=v-u/t

=23-8/3

=5m/s hope you got your answer

Explanation:

8 0
3 years ago
Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t
Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\  T

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

5 0
3 years ago
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