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SIZIF [17.4K]
2 years ago
7

When a narrow laser beam passes through a fine wire mesh before arriving at the wall, it forms a complicated pattern of bright s

pots on the wall. This pattern of spots would not occur if you sent a flashlight beam through the mesh because light from the flashlight is not a single electromagnetic wave. cannot be sent through a single opening of the mesh. is horizontally polarized, while laser light is vertically polarized. is vertically polarized, while laser light is horizontally polarized.
Physics
1 answer:
ser-zykov [4K]2 years ago
7 0

Answer:

this pattern to occur there must be coherence in the light beams.

you use a flashlight, the rays are incoherent so diffraction patterns cannot occur.

Explanation:

The point pattern that appears in the wall is the result of the interference and diffraction processes through each space of the mesh, for this pattern to occur there must be coherence in the light beams.

The coherence process is that all the rays have the same constant and phase, before the appearance of the lasers, the light is stopped by a small opening and this ray is the one that passes through the slits, with the appearance of the laser this it is consistent from its production process, so opening is not necessary, with this there is much greater intensity and the measurement process is simplified.

When you use a flashlight, the rays are incoherent so diffraction patterns cannot occur.

Polarization has no effect on diffraction patterns so it does not matter if it is vertical or horizontal.

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Answer:

When you ask a question, only two people can answer. When there are two answers, a little crown should appear at the bottom right hand corner. All you have to do is click the crown and it gives Brainliest. But you can only give it to one person per question

Explanation:

8 0
3 years ago
Which statement is true about the technology used to improve the motion of space vehicles on rough surfaces?
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The correct answer is C.)

It has made road vehicles safer because magnetometers are used to detect particles found in radiation emitted during combustion of fuel.

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3 years ago
A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.
jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m  </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

4 0
3 years ago
A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o
Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

W=mgh

W=32\times9.8\times5

W=1568\ J

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the  work done on the water.

\Delta U=W

\Delta U=1568\ J

The  change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

\Delta U=mc_{p}\Delta T

\Delta U=mc_{p}(T_{2}-T_{1})

T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}

T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}

T_{2}=23.01^{\circ}\ C

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0
2 years ago
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Artemon [7]

Answer:

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Explanation:

6 0
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