1. [2pt] The wind blowing across the surface of a pond creates ripples with a wa
ID: 1516111 • Letter: 1
Question
1. [2pt]
The wind blowing across the surface of a pond creates ripples with a wavelength of 7.20 cm. If these ripples propagate with a speed of 2.00 m/s, what is their period?
Answer:
2. [2pt]
The refractive index of a transparent material can be determined by measuring the critical angle when the material is in air (nair = 1.00). If c = 41.5 deg, what is the index of refraction of the material?
Answer:
3. [2pt]
For the previous problem, what is the speed of the light in the plastic?
Answer:
4. [2pt]
The temperature of a sample is measured to be 180.0 deg. F. What is this temperature expressed in Celsius (degC)?
Answer:
5. [2pt]
What is the temperature in the previous problem expressed in Kelvin (K)?
Answer:
Explanation / Answer
Ans:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22kAns:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22kAns:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22kAns:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22kAns:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22kAns:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22kAns:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22k
Ans:-
1 = 7.20cm = 0.072m v= 2m/s
T= /v = 0.072/2 =0.036s
2.by snell’s law
N1sin1 = n2sin2
1*sin41.5 =n2 sin90
n2 = 0.66
3. the speed of the light in the plastic
=3*10^8/0.66 = 4.55*10^8m/s
4 temperature = 180degF
TF = 9/5 Tc +32
180 = 1.8 Tc +32
Tc = 82.22c
5 Tk = 273 + Tc = 273+82.22 = 355.22k
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