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AlladinOne [14]
2 years ago
13

Which statement is true regarding copper's ability to conduct electricity? Select one: A. Copper is a good conductor of electric

ity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms. B. Copper is a poor conductor of electricity because its atoms are held in positions that cannot move. C. Copper is a good conductor of electricity because its atoms have electrons that are tightly bound to their shells and will resist movement. D. Copper is a poor conductor of electricity because it has a free electron in its outer shell that will flow to other copper atoms.
Physics
2 answers:
Elan Coil [88]2 years ago
5 0
<span>A.  Copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms.</span>
lidiya [134]2 years ago
5 0

Answer: Option (A) is the correct answer.

Explanation:

Copper is a transition metal and its atomic number is 29. The electrons in its shell are distributed as 2, 8, 18, 1.

As there is only one electron in the outer most shell and as it is far away from the nucleus, therefore, it is a loosely held electron. Also being a metal, copper is a good conductor of heat and electricity.

Thus, we can conclude that out of the given options the statement copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms is true regarding copper's ability to conduct electricity.

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How does the wavelength of a wave change when frequency decreases? when frequency increases?
Law Incorporation [45]

When the frequency decreases the wavelength is further apart. When it increases its closer together. Think about a flat line when the frequency is low the wavelengths are wider. When its a high frequency the squiggly lines on the moniter are taller and thinner so the wavelengths are not as wide and not that far from each other depending on how high the frequency is.

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3 years ago
A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.
Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

  1. Under the assumption that the can is a closed system, the conservation law applied to the system would be: E_{in}-E_{out}=E_{change}, where E_{in} is all energy entering the system, E_{out} is the total energy leaving the system and, E_{change} is the change of energy of the system.
  2. As the purpose is to kept the beverage can at constant temperature, the change of energy (E_{change}) would be 0.
  3. The energy  that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4) where \varepsilon is the emissivity of the surface, \sigma=5.67*10^{-8}\frac{W}{m^2K} known as the Stefan–Boltzmann constant, A_S is the total area of the exposed surface, T_S is the temperature of the surface in Kelvin, T_{\infty} is the environment temperature in Kelvin.
  4. For the can the surface area would be ta sum of the top and the sides. The area of the top would be A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2, the area of the sides would be A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2. Then the total area would be A_{total}=A_{top}+A_{sides}=0.01624m^2
  5. Then the radiation heat transferred to the can would be Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W.
  6. The can would lost heat evaporating water, in this case would be Q_{out}=\frac{dm}{dt}*h_{fg}, where \frac{dm}{dt} is the rate of mass of water evaporated and, h_{fg} is the heat of vaporization of the water (2257\frac{J}{g}).
  7. Then in the conservation balance: Q_{in}-Q_{out}=Q_{change}, it would be1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0.
  8. Recall that 1W=1\frac{J}{s}, then solving for \frac{dm}{dt}:\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}
5 0
2 years ago
A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image dista
KonstantinChe [14]

Answer:

-353.16

Explanation:

4 0
2 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
What happens to a satellite that slows down?
Bess [88]

Two things can happen to <u>old satellites</u>: For the closer satellites, engineers will use its last bit of fuel to slow it down so it will fall out of orbit and burn up in the atmosphere. Further satellites are instead sent even farther away from Earth.

4 0
3 years ago
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