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I really need someone to help me with these physics math problems

  • #1
Since no one seems to be helping me, maybe you will help me when I post my exact work that I did. I just need someone to check them all and make sure I did the right equations and if I got the right answers or not. This is for a grade, and I desperately need a passing grade on this worksheet to even pass the class. I made an honest attempt at each.


analysis:
10. the picture that we see in a mirror is called the
my answer=image
11. the actual object that makes a picture that we see on a mirror is called the
my answer=specular

-problems- show your work

15. What is the wavelength of a wave with a frequency of 100.4 MHz?

Given:
f=100.4 MHz
3x10^8

f=100.4x10^6

1004000000
-----------
3x10^8

=.335 m

16. A substance allows light to travel through it with a velocity of 2.45 x10^8 m/s. What is the substances index of refraction?

Given:
v=2.45x10^8 m/s
3x10^8

n=c/v

2.45 x 10^8
------------
3 x 10^8

=1.22

17. What is the intensity of a light that has a lumnious flux of 750 lumens?

Given:
P= 750 lm
E= ?

E=P/4(pie)

E=750 lm/4(pie)

E=589.05 cd

18. What is the illuminance provided by a light with a luminous flux of 1120 lm when it is 2.3 meters away?

Given:
E= ?
P= 1120 lm
d= 2m

E=P/4(pie)d^2

E=1120 lm/4(pie)2^2

E=22.28 lx

19. What would the angle of refraction be if a light ray starts in a diamond and goes into the air it has an angle of incidence of 9 degrees.

Given:
airn=1
n=?
90 degrees

n1sinQ1=n2sinQ2

sin(9)=n2sin(nr)
2.417xsinQ2=1sin90

=24.44 degrees

20. What is the critical angle of light going from quartz to air?

Given:
nquartz=1.544

sin x= 1/1.544 = .647

sin-1(.647)

=40.37 degrees

22. What is the illuminance of a light when it is 4.6 m away from a light with an intensity of 78 candle power?

d= 4.6 m
cd= 78
P= 980.2

E=P/4(pie)d2
cd=P/4(pie)

CD4(pie)=P
78(4)(pie)

E=980.2/4(pie)(4.6)^2

E=3.69 lx
 

Answers and Replies

  • #2
I post my work and I still get no help. I have asked politely and everyone refuses to help me. I don't get it.
 
  • #3
20
0
I would help if I could... no one seems to ever reply to my questions either. One time I even went as far as scanning my paper with what I had done so far and no one replied.
 
  • #4
dextercioby
Science Advisor
Homework Helper
Insights Author
12,977
540
Check the arithmetics in #17.

Daniel.
 
  • #5
What'd you see wrong in 17?
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
12,977
540
Isn't one candela one lumen in one sterradian...?

Daniel.
 
  • #7
I don't think so
 
  • #9
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
35
indiansfan4life said:
analysis:
10. the picture that we see in a mirror is called the
my answer=image
11. the actual object that makes a picture that we see on a mirror is called the
my answer=specular
The answer to 10 sounds reasonable enough. But your answer to 11 doesn't make a lot of sense to me. As far as I know, "specular" is a description, not a thing. It describes reflection from a smooth, polished surface, if I remember right. That having been said, I have no idea what #11 could be asking for, especially since they already used the word "object".

indiansfan4life said:
-problems- show your work

15. What is the wavelength of a wave with a frequency of 100.4 MHz?

Given:
f=100.4 MHz
3x10^8

f=100.4x10^6

1004000000
-----------
3x10^8

=.335 m

16. A substance allows light to travel through it with a velocity of 2.45 x10^8 m/s. What is the substances index of refraction?

Given:
v=2.45x10^8 m/s
3x10^8

n=c/v

2.45 x 10^8
------------
3 x 10^8

=1.22
15. [tex] c = f \lambda [/tex]. Solve for lamda. I don't your answer is correct. Recheck your calculations and algebra.

16. Your formula for the index of refraction is correct. *Note that you have reversed the numerator and denominator after plugging the numbers in*. Nevertheless, I'm assuming you did it correctly on your worksheet, since your answer is > 1.

Sorry...I don't remember enought about lumens, candela, lux and all those other crazy units to help you there.

Note: [itex] \pi [/itex] is spelled, "pi".


indiansfan4life said:
19. What would the angle of refraction be if a light ray starts in a diamond and goes into the air it has an angle of incidence of 9 degrees.

Given:
airn=1
n=?
90 degrees <---- what does this mean?

n1sinQ1=n2sinQ2

sin(9)=n2sin(nr) <---- what does nr mean?
2.417xsinQ2=1sin90 <---- where did 90 come from ?

=24.44 degrees
I think you've got the basic idea, *except that we are going from diamond to air, not the other way around!*. So let air be medium 2, and diamond be medium 1. Applying Snell's law, it should be:

[tex] n_1 \sin \theta_1 = n_2 \sin \theta_2 [/tex]

n2 = 1 for air.
n1 = ? (I'm assuming 2.417, since that is what you wrote)

[tex] \theta_2 = \sin^{-1} (n_1 \sin(9^o)) [/tex]


indiansfan4life said:
20. What is the critical angle of light going from quartz to air?

Given:
nquartz=1.544

sin x= 1/1.544 = .647

sin-1(.647)

=40.37 degrees
I think you did this one correctly. Key concept: at the critical angle of incidence, the angle of refraction is 90 degrees. The sine of 90 degrees is 1.
 

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