Answer:
Compared to windshield the airbag exerts much lesser force
Explanation:
Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time
<em>Impulse = Impulsive force *time</em>
I = F*Δt
If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.
Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.
During the collision, the passenger is carried towards the<em> windshield</em> and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.
The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.
<em>Thus compared to the windshield the airbag exerts much lesser force.</em>
<em> </em>
5.4 x 1014Hz
wavelength x frequency = the speed of light
Answer:
θ_p = 53.0º
Explanation:
For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º
Let's write the transmission equation
n1 sin θ₁ = ne sin θ₂
The angle to normal (vertcal) is
180 = θ2 + 90 + θ_p
θ₂ = 90 - θ_p
Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray
We replace
n1 sin θ_p = n2 sin (90 - θ_p)
Let's use the trigonometry relationship
Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p
In the law of reflection incident angle equals reflected angle,
ni sin θ_p = ns cos θ_p
n₂ / n₁ = sin θ_p / cos θ_p
n₂ / n₁ = tan θ_p
θ_p = tan⁻¹ (n₂ / n₁)
Now we can calculate it
The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water
n₂ = 1.33
θ_p = tan⁻¹ (1.33 / 1)
θ_p = 53.0º
n₂ = 1.40
θ_p = tan⁻¹ (1.40 / 1)
Tep = 54.5º
