#18**+10 **

*... what is the "electrical resistance nature" ... if there is additional resistance than we calc the current will be lower !*

There is not additional resistance. In this case, there is actually **less** resistance, so the current is higher.

The "electrical resistance nature" is simply the phenomena that results from the physical properties of any material to a change of temperature. In this case, it is resistance. The values are consistent with tungsten. My comments were to explain why the calculated value is different from the closest selectable value on the test.

All materials have a certain specific resistance and they change resistance according to temperature by certain amounts. Most (metal) materials have a positive resistance coefficient (negative conductance coefficient) to temperature. This is generally known as the "alpha" or “sigma” constant. The positive coefficient means the material increases in resistance to current flow as the temperature increases. For a negative coefficient the material decreases in resistance to current flow, as the temperature increases.

(The inverse of resistivity is called conductivity. There are perspectives where the use of conductivity is more suitable).

This coefficient is a hypothetical constant --though it really is not linear, meaning the rate of resistance to a given temperature will increase faster than the rate of temperature change. (Some materials become extremely nonlinear such as super conducting materials). However, for small (and often large) changes in temperature, the rate of change is usually insignificant, and a standardized value is used for basic calculations.

For tungsten the alpha" (α) constant is 0.004403 ohms per degree C. The initial reference point is 5.6E-8 ohms per meter @ 20 C. This means an increase of 227C will increase the resistance by 1 ohm uniformly across the standardized area and length.

Tungsten filaments are designed to operate near the lower end of its vaporization point to efficiently produce the greatest light per unit of energy. This is approximately 2500 to 2800 Kelvins. When a bulb is operated at a lower voltage the energy is reduced and the operation temperature drops accordingly. The temperature drop will in turn lower the resistance. The lower resistance will increase the current flow in proportion. Keep in mind, too, that as the resistance drops and the current flow increases the power also increases. At some point this reaches an equilibrium.

In the example, assume the calculated and actual current flow is different by about 0.7 amp, this would correspond to a temperature reduction of 159C per standard unit. Most (higher wattage) bulbs have a coiled filament of about 20 to 22 cm (the cross sectional area is also smaller). Without knowing accurate specifications, the calculations are somewhat arbitrary, but two to three times this temperature is reasonable.

The concepts presented here are usually covered in more detail in the advanced physics classes. The phenomena is described in part at the quantum level, how electron motion is affected and mediated by short-range spin waves known as paramagnons.

A basic analogy to describe the effect: Consider how a person might navigate across a room full of people. If the people are standing relatively still then navigating is more efficient than if the people are randomly moving around. The heat in this case corresponds to the motion of the people moving around.

~~D~~

DavidQD Aug 15, 2014

#1**+10 **

Most of them are, but:

16. Metal feels colder than wood because the rate it draws heat from your hand is greater than that of wood. Because it's the *rate of transfer of energy* that is important here, it is the specific heat capacity (larger for wood than metal, so rate of change of temperature is smaller) that matters most.

25. Red light is refracted least.

30. Atoms radiate photons when they undergo a transition from a high energy state to a low one.

I can't read question 7 in the last image!

Alan Aug 14, 2014

#2**+10 **

The bulb when operated at rated voltage of 240V uses 280W per hour. This indicates a resistance of

E^{2}/P (240^{2})/280W = 206 ohms.

Assuming a uniform resistance, the current (I) at 200V (E), with a resistance of 206 ohms equals * E/R =I -- - -> 200/206 = =0.972* The correct answer is 0.98 (b)

Note: the actual resistance is slightly lower than the calculation. The electrical resistance nature of tungsten based filaments is non-liner, and produces a slightly lower resistance at lower temperatures. The lower resistance results in a higher current for the same voltage. This is probably why the answer indicates 0.98 instead of 0.97.

---------

For # 2 Your answer is correct.

From the Lorentz Force Law, for parallel motion in a magnetic field, the sine of the relative angle is zero.

---------

For # 12 Your answer is correct.

A= F/m * where, m = Mass, a = acceleration, and F= force

---------

For # 13 Your answer is correct.

The two temperatures scales converge at -40

---------

For # 16 Your answer is correct.

---------

For # 17 Your answer is correct.

---------

For # 18 Your answer is correct.

The value is very small: 8.17E-10 N

---------

For # 19 Your answer is correct.

---------

For # 20 Your answer is correct.

---------

For # 22- #24 Your answers are correct.

---------

For # 25 The answer is (A). Red light is the least refracted.

---------

For # 26 Your answer is correct

---------

For # 29 Your answer is correct. (Though the magnitude is the same the vector is opposite).

---------

For # 30 The answer is (A).

---------

The top part of the last image is indiscernible.

The lower portion.* For # 8. I am uncertain.

Transverse waves in matter are a displacement of the medium that is perpendicular to the direction of propagation of the wave.

Most graphs depict Amplitude in vertical format.

If this is the case here then the answer is (D) for D and E. For peak to peak amplitude.

If this is an exception to the standard (because of the waves nature) then (C) for A and C, is correct.

(Alan may be able to clarify this).

---------

~~D~~

DavidQD Aug 14, 2014

#3**+5 **

In the last part of the last image it looks to me as though *B) A and E* is circled. This is correct here, as amplitude is usually mean-to-peak. Think of the curve as y = a*sin(x). a is generally thought of as the amplitude and y varies from +a to -a.

NB David is correct about question1. I missed that!

Alan Aug 14, 2014

#4**+10 **

Thanks Alan. The “mean-to-peak” for amplitude is helpful. (I was uncertain what applied to transverse waves).

----

Interesting point on # 16.

I considered the specific heat aspect, but thought the psychometric formulas to have more weight here. Consider two substances that have a similar specific heat, but different conduction coefficients. Basically applying the same amount of heat will elevate each temperature the same amount. The rate at which these will absorb heat are different.

Though the total energy transferred for a given temperature rise is the same, the psychometric formulas would indicate a much colder “feel” by touch because the energy transfer is faster.

(I have never seen a collegiate physics test that didn’t have at least one question with ambiguity -- “The ubiquitous trick question.”)

~~D~~

DavidQD Aug 14, 2014

#5**+10 **

Re. 16). You might well be right David. I just thought in terms of a relatively small block of metal/wood for which one might write m.c_{p}.dT/dt = heff.A(T_{hand} - T) where heff is an effective heat transfer coefficient that would include thermal conductivity as well as a contact heat transfer coefficient. In this case the heat transfer is likely to be dominated by c_{p} when it transfers to the right-hand side.

If we are talking about large blocks, with effectively infinite heat capacity, then it is probably better to write a similar equation from the hand's point of view, in which case it will probably be the contact resistance that dominates.

Never really thought about it much before - an interesting question to ponder!

Alan Aug 14, 2014

#7**+10 **

xvxvxv: *In question number 1 why I can't use P=V.I*

You can, but you have to do it at the normal condition first: so 280=240*I_{normal}

I_{normal} = 280/240.

Ohms law says V = IR so 240 = R*280/240 or R = 240^{2}/280

The resistance R, will be approximately the same at reduced voltage, so using Ohm's law again:

200 = I_{reduced}*240^{2}/280

I_{reduced }= 200*280/240^{2}=0.972

(PS on question 16, go with thermal conductivity!).

Alan Aug 14, 2014

#8**0 **

I can't understand it !

Now the bulb operated at 200 v and it power is 280

so I can find the current by I=P/V ..... I=280/200 ... I = 1.4 A

I want to understand my mistake in this way !

also in question 25 ! why the red ..؟

I'm based on this law dsinx=mL x :- is the angle and L :- is the wave length ! ?

xvxvxv Aug 14, 2014

#9**+10 **

*Now the bulb operated at 200 v and it power is 280*

---

No, it did not.

The bulb operates at 280W when a potential of 240V is its source.

When the voltage is lowered to 200v the energy used is lowered in proportion. In this case the bulb is using 192 watt-hours, not 280wh.

(Note error: The 192 watt-hours should be **194 **watt-hours)

Resistance is hypothetically constant. It is determined using the rated standard operating conditions. The power is known and the voltage is known. Squaring the voltage then dividing by the power yields resistance. Dividing the (now) known resistance into the 200v source yields the current (in amps). If you then take the product of the amperage (0.97) and applied source voltage(200v) this yields the power (194 watts).

Remember the bulb only uses 280 watts when the voltage is 240V.

~~D~~

DavidQD Aug 14, 2014

#12**+10 **

*you mean 194 watt not 192 ... right ? *

You are correct. (The value depicted is a typographical error) A correction is noted.

---

*what about the refractive !! why the red is the least *

One (sometimes confusing) point to keep in mind, is the difference between **diffraction** and **refraction**.

Longer wavelengths –near the red zone -- are **diffracted** more, but they **refract** less. The opposite is true for shorter wavelengths.

~~D~~

DavidQD Aug 14, 2014

#16**0 **

David

in your first answer you write

Note: the actual resistance is slightly lower than the calculation. The electrical resistance nature of tungsten based filaments is non-liner, and produces a slightly lower resistance at lower temperatures. The lower resistance results in a higher current for the same voltage. This is probably why the answer indicates 0.98 instead of 0.97

what is the **"electrical resistance nature" ... **if there is additional resistance than we calc the current will be lower !

xvxvxv Aug 15, 2014

#18**+10 **

Best Answer

*... what is the "electrical resistance nature" ... if there is additional resistance than we calc the current will be lower !*

There is not additional resistance. In this case, there is actually **less** resistance, so the current is higher.

The "electrical resistance nature" is simply the phenomena that results from the physical properties of any material to a change of temperature. In this case, it is resistance. The values are consistent with tungsten. My comments were to explain why the calculated value is different from the closest selectable value on the test.

All materials have a certain specific resistance and they change resistance according to temperature by certain amounts. Most (metal) materials have a positive resistance coefficient (negative conductance coefficient) to temperature. This is generally known as the "alpha" or “sigma” constant. The positive coefficient means the material increases in resistance to current flow as the temperature increases. For a negative coefficient the material decreases in resistance to current flow, as the temperature increases.

(The inverse of resistivity is called conductivity. There are perspectives where the use of conductivity is more suitable).

This coefficient is a hypothetical constant --though it really is not linear, meaning the rate of resistance to a given temperature will increase faster than the rate of temperature change. (Some materials become extremely nonlinear such as super conducting materials). However, for small (and often large) changes in temperature, the rate of change is usually insignificant, and a standardized value is used for basic calculations.

For tungsten the alpha" (α) constant is 0.004403 ohms per degree C. The initial reference point is 5.6E-8 ohms per meter @ 20 C. This means an increase of 227C will increase the resistance by 1 ohm uniformly across the standardized area and length.

Tungsten filaments are designed to operate near the lower end of its vaporization point to efficiently produce the greatest light per unit of energy. This is approximately 2500 to 2800 Kelvins. When a bulb is operated at a lower voltage the energy is reduced and the operation temperature drops accordingly. The temperature drop will in turn lower the resistance. The lower resistance will increase the current flow in proportion. Keep in mind, too, that as the resistance drops and the current flow increases the power also increases. At some point this reaches an equilibrium.

In the example, assume the calculated and actual current flow is different by about 0.7 amp, this would correspond to a temperature reduction of 159C per standard unit. Most (higher wattage) bulbs have a coiled filament of about 20 to 22 cm (the cross sectional area is also smaller). Without knowing accurate specifications, the calculations are somewhat arbitrary, but two to three times this temperature is reasonable.

The concepts presented here are usually covered in more detail in the advanced physics classes. The phenomena is described in part at the quantum level, how electron motion is affected and mediated by short-range spin waves known as paramagnons.

A basic analogy to describe the effect: Consider how a person might navigate across a room full of people. If the people are standing relatively still then navigating is more efficient than if the people are randomly moving around. The heat in this case corresponds to the motion of the people moving around.

~~D~~

DavidQD Aug 15, 2014

#19**0 **

Thanx david .. you are amaizng

now I want to ask about question 16 .. I saw that you have long discuss with alan .. and the English is not my mother language .. so I want some one explain to me the answer with simply .. why it is not specific heat capacity ...

xvxvxv Aug 16, 2014

#20**+5 **

The answer to question 16 isn't, perhaps, as simple as the person who set the question might have thought. Here are some links that discuss the problem; some like thermal conductivity, some like specific heat!!

https://uk.answers.yahoo.com/question/index?qid=20120514131608AAAhh9l

http://boards.straightdope.com/sdmb/showthread.php?t=143186

http://classroom.synonym.com/steel-feel-colder-wood-5918.html

Alan Aug 16, 2014

#22**+5 **

The principle consideration here is definition of the word “feel” used in the test question. This is a subjective quality and is open to a broad range of interpretations.

When it comes to physics and other physical sciences, there is often, if not always, a need to quantify uniformly (as much as possible) a subjective quality. For example, why does a person feel hot at times then cold at other times when subjected to the same temperature?

To give a basis for understanding and eventually controlling environments, a series of equations were developed to quantify environmental effects on living beings (including animals). These equations are generally referred to as psychometric equations. The most recognized of these are the wind-chill factor and Heat index (aka humiture or humidex). Though common in usage, most do not really understand the information these equations present.

Other psychometric equations include the perception of sound and light. Though, not necessarily a psychometric equation, the decibel is used for sound energy because of the way the human ear perceives sound.

Most psychometric equations represent a rate. In the case of the wind-chill factor, it correlates the equivalent rate at which heat is removed by wind to a temperature without wind. A similar relation exists for the heat index. Though theses give values, the equation relates how a person perceives an effect relative to another effect.

The case in point for this question has more to do with how fast heat is removed from a human hand than how fast a substance change temperature for a given amount of heat. Though these two are related, they are not really the same thing.

~~D~~

DavidQD Aug 16, 2014

#23**+5 **

I agree that "feel" is somewhat subjective; but, in this case, most people would agree on the effect (I've just tried it myself and it's fairly clear-cut) if not the magnitude of it. To me, that suggests there is an underlying physical mechanism, related to the material being touched. My further thoughts on this are as follows:

There are ** two** physical requirements for removal of heat from a finger, say, when it touches a solid surface.

The ** first** is that the material in contact with the finger is capable of conducting the heat away. A better conductor will conduct the heat away better, so one up for thermal conductivity.

However, the ** second** requirement is that there be a temperature difference between the material in contact with the finger and the rest of the material. If the local temperature of the solid doesn't rise, the heat will not flow away no matter how good the thermal conductivity. The local material will rise in temperature quicker and hence start to remove heat more quickly if it has a smaller specific heat capacity. One up for specific heat capacity.

Both properties are involved.

I'm tempted to construct a simple mathematical model of the process!

Alan Aug 16, 2014