An ∃mazing Physics thread (Unit 2)

[Guest]

Hi everyone,

I shall start a Physics questions thread now mainly for myself as I see that there is already a 3&4 thread. Hopefully others will benefit from this as well.

Question:
1) Sound travels with a speed of 343m/s in air. Find the speed of sound in steel where K=200 GPa and p= 7,870 kg/m^3, and in water where E=2,200 GPa and p=1000 kg/m^3. Compare to the speed in air.

Normally I can solve these questions by simply plugging the values into the equations but in this case the value of K is provided for a solid object and the value of E is given for the liquid. Shouldn't it be the other way around? Would some conversion be necessary? How would you "compare to the speed in air"?

The answer states that : the speed is 5,040 m/s in steel and 4.7*10^4 m/s in water. Could someone please explain how one might reach this conclusion?

Thank You

[Yacoubb]

Hi everyone,

I shall start a Physics questions thread now mainly for myself as I see that there is already a 3&4 thread. Hopefully others will benefit from this as well.

Question:
1) Sound travels with a speed of 343m/s in air. Find the speed of sound in steel where K=200 GPa and p= 7,870 kg/m^3, and in water where E=2,200 GPa and p=1000 kg/m^3. Compare to the speed in air.

Normally I can solve these questions by simply plugging the values into the equations but in this case the value of K is provided for a solid object and the value of E is given for the liquid. Shouldn't it be the other way around? Would some conversion be necessary? How would you "compare to the speed in air"?

The answer states that : the speed is 5,040 m/s in steel and 4.7*10^4 m/s in water. Could someone please explain how one might reach this conclusion?

Thank You

Is this even a 1+2 question lol?

[Guest]

Is this even a 1+2 question lol?

Hahaha. It is.

[Yacoubb]

Hahaha. It is.

Unit 2?
Haha well at my school we've only started looking at speed, velocity, looking at displacement-time and velocity-time graphs and calculating values from there. Never looked at sound.. yet!

[lzxnl]

Hi everyone,

I shall start a Physics questions thread now mainly for myself as I see that there is already a 3&4 thread. Hopefully others will benefit from this as well.

Question:
1) Sound travels with a speed of 343m/s in air. Find the speed of sound in steel where K=200 GPa and p= 7,870 kg/m^3, and in water where E=2,200 GPa and p=1000 kg/m^3. Compare to the speed in air.

Normally I can solve these questions by simply plugging the values into the equations but in this case the value of K is provided for a solid object and the value of E is given for the liquid. Shouldn't it be the other way around? Would some conversion be necessary? How would you "compare to the speed in air"?

The answer states that : the speed is 5,040 m/s in steel and 4.7*10^4 m/s in water. Could someone please explain how one might reach this conclusion?

Thank You

Ok firstly, if you're looking at this in year eleven, you have a slightly strange curriculum.
Secondly, for this question, just plugging the values into the formula would give you answers that match the answers given. A quick online search shows that the "E" value given is actually the K value for water.

[Guest]

Ok firstly, if you're looking at this in year eleven, you have a slightly strange curriculum.
Secondly, for this question, just plugging the values into the formula would give you answers that match the answers given. A quick online search shows that the "E" value given is actually the K value for water.

Yes, that's exactly what I thought. Perhaps the book made a mistake in defining the values?

[götze]

LoL he goes to johns monash what else do you accept  ?

[Guest]

I don't seem to be understanding the "link" between two questions regarding 'total internal reflection'.

Question 1: Can a light ray propagating from air to water suffer total internal reflection at any angle?
Answer: No, it cannot. I found the answer quite easily by applying the values of refraction in water (1.33) and air (~1). Snell's Law tells us that: sin^-1(1.33) does not have any solutions, meaning that there is no total internal reflection at any angle.

Question 2: A light source is placed at the bottom of a swimming pool, and a light ray directed toward the surface of water makes an incident angle of 68 with the normal. Find the angle of refraction in air.
Answer: I solved this using the same method for question one since it is important to check for total internal reflection first. Snell's Law now tells us that 1.33=1*sin(x). This is where I get stuck. The book answers say that the light is totally internally reflected. I thought it would not be totally internally reflected at all because that's exactly what happened in the previous question (Question 1)

My question:What is the relation of "sine" to the light that's totally internally reflected?     

[lzxnl]

You have the situations confused. The first case is from air to water, from lower refractive index to higher.
The second case is from water to air, from higher refractive index to lower.
The two situations are not identical. TIR only occurs when the refracted medium has a lower refractive index.

[Guest]

Should be an easy one:

Calculate the critical angle for light travelling through a diamond (n=2.5) towards the surface.

I keep making some silly mistake 

Thanks.

[Kuroyuki]

Should be an easy one:

Calculate the critical angle for light travelling through a diamond (n=2.5) towards the surface.

I keep making some silly mistake 

Thanks.

Im pretty sure the formula is n = 1/sin(critical angle )
So 2.5 = 1/sin I
1/2.5 = sin I
0.4 = Sin I
So the incident angle comes to 23.57 degrees.
You can verify using smells law.
1sin90=2.5sin(23.57)
Hope this helps and hope this isn't wrong. Physics isn't my strongest subject.

[Guest]

Thanks, I just worked it out before you replied. However, I really like how you verified using Snell's Law as well. I guess that technique will come in handy some time. 

[Guest]

This question is slightly ambiguous in the sense that the values for refractive index of the corresponding substance (Perspex) is not stated.

Question: What is the relative refractive index for light passing from Perspex into water if an incident angle of 17.0 produces an angle of refraction of 14.5?

I would be able to solve if the refractive index of one of the substances was stated or by referring to the table of refractive index values, but they are not specifically provided in this case. Also, in which substance is the question asking to find the relative refractive index for? I'm guessing its water.

Thanks 

[09Ti08]

I think you're confused between "refractive index" and "relative refractive index". You should pay attention to the word "relative" in Physics.

If you have a look at the definition of refractive index: n=c/v, then it tells you that the refractive index is defined based on the speed of light in a medium with respect to that in vacuum.

Now, relative refractive index of medium 1 with respect to medium 2 is different: n12=c1/c2. In this case, the relative refractive index is defined based on the speed of light in medium 1 with respect to medium 2. If you want, you can think that refractive index is obtained by letting medium 2 to be vacuum.

Ok, so: n12=c1/c2=n2/n1=sin(theta 1)/sin(theta 2)=sin 17/sin 14.5=1.17

[Guest]

Been a while since I posted for Physics help since this Motion chapter has been excruciatingly long. I've asked many this question before and it has been approached in different ways every time. I would like to know how people go about solving it in the most succinct but complete way.
Thanks.

A force of 120 N is used to push a 20 kg shopping trolley along
the line of its handle—at 20° down from the horizontal. This is
enough to cause the trolley to travel with constant velocity to
the north along a horizontal path.

a) What is the value of the frictional force acting against the
trolley?
b) How large is the normal force that is supplied by the ground
on which the trolley is pushed?