Answer:
a) wavelengths seen decrease
, b) λ’/λ = 0.9998
Explanation:
a) This is a relativistic doppler effect problem that is described by
f’= f₀ √ (1+ v / c) / √ (1- v / c)
The speed of light be related to wavelength and frequency
c = λ f
f = c /λ
We replace
c /λ’= c /λ √ (1+ v / c) / √(1- v / c)
λ’= λ √ [(1-v / c) / (1 + v / c)]
This expression gives us the wavelength depends on the speed of the premiere with respect to the Earth, as the estuary approaches the Earth the speed is positive
We can see that at the root it is less than 1 whereby the wavelengths seen decrease
b) we reduce the speed to m / s
v = 30 km / s = 3 10⁴ m / s
Change is
λ’/λ = √ [(1-v / c) / (1 + v / c)]
λ’/ λ = √ [(1- 3 10⁴/3 10⁸) / (1 + 3 10⁴/3 10⁸)]
λ’/ λ = √ (0.9999 / 1.0001)
λ’/λ = 0.9998
Answer:
If conditions are just right, you can see Polaris from just south of the equator. Although Polaris is also known as the North Star, it doesn't lie precisely above Earth's North Pole. If it did, Polaris would have a declination of exactly 90 degree.
Explanation:
Voltage = current * resistance
V = I*R
I = 4 A
R = 2 ohm
V = 4 A * 2 ohm
= 8 volt
Answer:
The index of refraction of the first medium must be higher than the index of refraction of the second medium
Explanation:
Snell's law describes the behaviour of light at the boundary between two mediums:
where
n1 and n2 are the index of refraction of the two mediums
are the angle between the direction of the light ray and the normal to the interface
We can rewrite the condition as:
Let's assume now that the light is travelling in the first medium with a very large angle with respect to the normal to the surface, i.e. , so that . In this case, we have
We notice that if , the ratio on the right is larger than 1, and so the term should be also larger than 1: but this is not possible of course, since the sine function is always less than 1. Therefore, in this case total internal reflection occurs, because no refracted ray is produced.
Answer:
7650 m.
Explanation:
Ocean floor depth, d is:
d = v * t,
where,
d = the distance from the vessel to the ocean floor (or the depth)
v = 1530 m/s = velocity of the ultrasonic sound
t = t_echo/2 = time that the ultrasonic sound needs to reach the ocean floor
t_echo = 10 s = time that the ultrasonic sound needs to reach the ocean floor and return back to the vessel.
d = v * t
= v * t_echo/2
= 1530 * 10/2
= 7650 m.