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andre [41]
3 years ago
14

A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material suc

h as rubber is a(n) _______________ because it _______________ the flow of electric charges. A material that partially conducts electric current is a(n) _______________. These materials include _______________ elements.
Physics
2 answers:
notka56 [123]3 years ago
6 0

Explanation:

A metal such as copper is a <u>conductor</u> because it provides a pathway for electric charges to move easily. A material such as rubber is an <u>insulator</u> because it <u>resists</u> the flow of electric charges. A material that partially conducts electric current is a <u>semiconductor</u>. These materials include <u>group 3 and group 5</u> elements.

jeka57 [31]3 years ago
4 0

Answer:

conductor

insulator

resists

semiconductor

group 3 and group 5

Explanation:

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A parallel circuit has a 125 Volt battery connected with 3 resistors. R1= 20 Ω, R2= 100 Ω, and R3= 50 Ω. Find the total current
Musya8 [376]

Answer:

10A

Explanation:

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3 years ago
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You lift a 50 N object 2 meters off the ground what work did you do on the object
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Which type of wave interaction is shown in the diagram?
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3 years ago
An open pipe of length 0.58 m vibrates in the third harmonic with a frequency of 939Hz. What is the speed of sound through the a
vaieri [72.5K]

363 m/s is the speed of sound through the air in the pipe.

Answer: Option B

<u>Explanation:</u>

The formula used to calculate the wavelength given as below,

      Wavelength (\lambda)=\frac{\text { wave velocity }(v)}{\text { frequency }(f)}

      v=\lambda \times f   --------> eq. 1

In power system, harmonics define by positive integers of the fundamental frequency. So the third order harmonic is a multiple of the third fundamental frequency. Each harmonic creates an additional node and an opposite node, as well as an additional half wave within the string.

If the number of waves in the circuit is known, the comparison between standing wavelength and circuit length can be calculated algebraically. The general expression for this given as,

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For first harmonic, n =1

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For second harmonic, n =2

         L=\frac{2 \lambda}{2}=\lambda

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         L=\frac{3 \lambda}{2}

         \lambda=\frac{2 L}{3}   -------> eq. 2

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