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morpeh [17]
3 years ago
10

A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00

0 m/s north. What is the planes’ acceleration from A to B?
Physics
1 answer:
Airida [17]3 years ago
7 0

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

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A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl
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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

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     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

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Let's use the trigonometry relationship

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In the law of reflection  incident angle equals reflected angle,  

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    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

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3 years ago
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