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Svetach [21]
3 years ago
12

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur

e. The width of the central maximum is 2.5 cm .What is the diameter (in mm) of the hole?
Physics
1 answer:
devlian [24]3 years ago
5 0

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

                                             2 y = 0.025 m

                                               y = 0.0125 m

For circular aperture

  sin \theta = 1.22\dfrac{\lambda}{d}

using small angle approximation

  \theta = \dfrac{y}{D}

now,

   \dfrac{y}{D} = 1.22\dfrac{\lambda}{d}

   y = 1.22\dfrac{\lambda\ D}{d}

   d = 1.22\dfrac{\lambda\ D}{y}

   d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}

         d =0.247 x 10⁻³ m

         d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

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The process that enhances some properties of an object at the expense of other properties is called:
lidiya [134]

Answer:

The answer is: letter a, pop-out effect.

Explanation:

The "pop-out effect" is a phenomenon which allows the person's precognitive processes to detect a<em> visual stimulus that is potentially the most meaningful one</em> in a person's spatial field of attention. The pop-up effect occurs when a person distinguishes one object from the rest.

For example, when a child chooses among pictures in different colors, it is common for the child to point at colored pictures rather than grayscale pictures. This is an example of a pop-out effect. <u>The properties of the colored pictures is more preferred by the child thus, causing him not to choose or mind the grayscale images.</u>

Thus, this explains the answer.

6 0
3 years ago
The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
3 years ago
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MrRa [10]
The answer is A vaporization 

5 0
3 years ago
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Korolek [52]

Explanation: It is because when a car is moving both the car and the driver is in inertia of motion. When a car is involved in collision it comes to a sudden stop and the car comes into inertia of rest whereas the person still in inertia of motion moves forward and might result in major injuries. But this can be prevented by wearing a seatbelt

Hope it helps :)

8 0
2 years ago
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