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forsale [732]
2 years ago
7

A thin layer of oil (n = 1.38) on the ocean surface for thousands of miles. As an inspector, you are flying over the ocean surfa

ce in an aircraft, looking at a section of option where the oil thickness is 563 nanometers thick. You look straight down at the ocean, at a time of day when the sun is directly above you. a) So, if constructive interference occurs (i.e. where the reflection is brightest) what is the wavelength(s) of visible light in the reflection? b) However, in order to get a close-up look at the oil slick, you put on scuba gear and go diving under the portion of ocean you had been observing from the plane. For which wavelength(s) of visible light yield the strongest transmitted intensity?

Physics
1 answer:
Eddi Din [679]2 years ago
7 0

Answer:

Explanation:

The picture attached shows all the necessary explanations

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A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

7 0
3 years ago
The formula of force is newton and​
Arlecino [84]

Answer:

The formula for calculating force is F= Mass × Acceleration

Explanation:

Newton is what force is measured in

8 0
3 years ago
A person is riding a motorized tricycle. They weigh 180kg and are moving at 3 m/s over a distance of 300 m. How much work is don
agasfer [191]

If I am to understand this question correctly this is what asks you:

If a person is riding a motorized tricycle how much work do they do?

You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any <em><u>motorized vehicle</u></em> other than the time it takes to get from one place to another.

So to answer this question I would say:

Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.

If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!

3 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
3 years ago
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