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postnew [5]
2 years ago
8

When light travels through a small hole, it appears to be an observer that the light spreads out, blurring the outline of the ho

le. Does this observation support the theory of light as a wave, or light being made of particles? Explain.
Physics
1 answer:
3241004551 [841]2 years ago
7 0

Answer:

support lights as a wave

Explanation:

In the model of light as a particle, the experimenter would expect to see one small hole of light emerging on the wall. However, as the light spreads out, it behaves much like a wave that diffracts when going through a small hole.

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A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge
Kamila [148]

Answer 1) The electric field at distance r from the thread is radial and has magnitude  

E = λ / (2 π ε° r)  

The electric field from the point charge usually is observed to follow coulomb's law:  

E = Q / (4  π ε° r^{2})  

Now, adding the two field vectors:  

E_{thread}  =  {2.5 / (22 π ε° X 0.07 ) ; 0}  

Answer 2) E_{q}  = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.  

The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

5 0
3 years ago
Select the correct answer.
scZoUnD [109]

Answer:

B. It is directly proportional to the source charge.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

Hence, the statement which is true of the electric field at a distance from the source charge is that it is directly proportional to the source charge.

7 0
2 years ago
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
Rocks are elastic too! Explain how elasticity and Harry Reid's elastic rebound theory lead to a better understanding of earthqua
ANEK [815]
After the great 1906 San Francisco earthquake, geolophysicistHarry Fielding Reid examined the displacement of the ground surface along the San Andreas Fault. He concluded that the quake must have been the result of the elastic reboundof the strain energy in the rocks on either side of the fault.
strain energy is 0. 5x force x (compression) X (compression)
There is a lot of force and a bit of compression when rocks squash up against other rocks causing earthquakes
7 0
3 years ago
A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b
netineya [11]

Answer:

The acceleration is  a =51945 \ km/h^2

Explanation:

From the question we are told that

   The lift up speed is  v  = 147 \  km/h

    The distance covered for the take off run is s =  208 m = 0.208 \ km

Generally from kinematic equation we have that

      v^2 = u^2 + 2as

Here u is the initial  speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So  

    147^2 = 0^2 + 2* a* 0.208

=>  a =51945 \ km/h^2

6 0
2 years ago
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