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Alenkinab [10]
1 year ago
7

You notice a band of yellow and an adjacent band of green in a soap bubble. The green band is coming from a region of the bubble

that is?
Physics
1 answer:
Ainat [17]1 year ago
3 0

The green band is coming from the thicker region of the bubble.

White light is made up of all colours, all wavelengths. If one of these colours is subtracted from white light (by interference, for instance) we see the complementary colour.

For example, if blue light is subtracted from white light, we see yellow. The skin of a bubble glistens with the complementary colours produced by interference.

If we were to look at a highly magnified portion of a soap bubble membrane, we would notice that light reflects off both the front (outside) and rear (inside) surfaces of the bubble, but the ray of light that reflects off the inside surface travels a longer distance than the ray which reflects from the outside surface.

When the rays recombine they can get "out of step" with each other and interfere. Given a certain thickness of the bubble wall, a certain wavelength will be cancelled and its complementary colour will be seen. Long wavelengths (red) need a thicker bubble wall to get out of step than short wavelengths (violet). When red is cancelled, it leaves a blue-green reflection.

As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is cancelled, leaving yellow. Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background.

Learn more about soap bubble membrane here : brainly.com/question/13032930

#SPJ4

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1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

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