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Hunter-Best [27]
3 years ago
5

Which statement about metals is true?

Physics
1 answer:
Gnom [1K]3 years ago
4 0

All metals are solid at (our) living temperatures except for mercury which is a liquid. Metals are shiny for a while at least when properly finished. Metals are good conductors of both heat and electricity. This is due to the excess electrons in their valence clouds that facilitate energy transfer.What are 5 characteristics of metal?

Metals are lustrous, malleable, ductile, good conductors of heat and electricity

Silver conducts electricity better than any other metal.

Brass is an alloy made from zinc and copper.

Pure gold is too soft for many things so most gold is combined with other metals to make it stronger. ...

World time periods are often listed by the metal used.

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A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?
garri49 [273]

Answer:

2 m/s^2

Explanation:

a = v^2/r

a = (10m/s)^2 / 50m

a = 2 m/s^2

Leave a like and mark brainliest if this helped

Leave a like and mark brainliest if this helped

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3 years ago
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The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.
vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of q. Assume that V is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

\text{Potential Energy} = q \cdot V.

Keep in mind that since both q and V might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

  • q = \rm -8.0\; \mu C; (that's -8.0 microjoules, which equals to \rm -8.0\times 10^{-6}\; J)
  • V = \rm 40\; V.

Hence its electrical potential energy:

\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J.

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this \rm -8.0\; \mu C negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0
3 years ago
What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
Which letter represents the location of the resister in this diagram? A B C D​
Elodia [21]

Answer:

A

Explanation:

7 0
3 years ago
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