Answer:
Q = 47.06 degrees
Explanation:
Given:
- The transmitted intensity I = 0.464 I_o
- Incident Intensity I = I_o
Find:
What angle should the principle axis make with respect to the incident polarization
Solution:
- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:
I = I_o * cos^2 (Q)
- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:
Q = cos ^-1 (sqrt (I / I_o))
- Plug the values in:
Q = cos^-1 ( sqrt (0.464))
Q = cos^-1 (0.6811754546)
Q = 47.06 degrees
Answer:
Frequency, f = 15 Hz
Explanation:
We have,
Speed of an ocean wave is 45 m/s
Wavelength of a wave is 3 m
It is required to find the frequency of an ocean wave.
Speed of a wave,
, f = frequency of ocean wave

So, the frequency of an ocean wave is 15 Hz.
Answer:
A.
Explanation:
Earth is composed of different layers and one layer moves over another due to differences in the densities.
According to the physics of density, a substance having less density floats over a higher density substance. The oceanic crust has more density than the continental crust that is why continental crust float over oceanic crust.
So in the given example, plate B is moving below the plate A, it means plate B is more dense than plate A because plate B is composed of oceanic crust
. <u>For example : continents float over the asthenosphere (a layer below the lithosphere).</u>
Hence, the correct answer is "A
".
Answer:
V₀ = 5.47 m/s
Explanation:
The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 3.04 m
θ = Launch Angle = 41.7°
V₀ = Minimum Launch Speed = ?
g = 9.81 m/s²
Therefore,
3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)
V₀² = 3.04 m/(0.10126 s²/m)
V₀ = √30.02 m²/s²
<u>V₀ = 5.47 m/s</u>
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