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Sholpan [36]
2 years ago
9

Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

s each needle is less than a millimeter thick. So, to find the thickness, she uses a needle to poke a hole in a piece of brown construction paper. Then, she positions a 640 nm laser pointer to shine through the hole and project a circular diffraction pattern on a wall 21.7 m away. She then uses her ruler to measure that the central bright circle is 14.2 cm in diameter. What diameter does Irina calculate for the needle?
Physics
1 answer:
Vsevolod [243]2 years ago
8 0

Answer:

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by d sin\Theta =1.22\lambda--------eqn 1

The angle through which the first minima is diffracted is given by tan\Theta =\frac{y_1}{D}---------eqn 2

As \Theta is very small so we can write sin\Theta =tan\Theta

So from eqn 1 and eqn 2 we can write

y_1=\frac{1.22\lambda D}{d}--------eqn 3

Here y_1 is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so y_1=\frac{14.7}{2}=7.35 \ cm

Now putting all these value in eqn 3

d=\frac{1.22\lambda D}{y_1}

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

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Why is a solution of 4% acetic acid in 95% ethanol used to wash the crude aldol-dehydration product?
raketka [301]

A cold acetic acid solution is used to wash the residue of the reagent in preparation of an aldol condensation product after vacuum filtration.  The main reason in washing with the acetic acid rinse is to neutralize any sodium hydroxide.

5 0
2 years ago
Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by
ankoles [38]

Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

A virtual image can not be formed on a screen.

For image:

1.A virtual image can be viewed by the unaided eye.

2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

So the option A is correct.

5 0
2 years ago
An aircraft on it's take-off run has a steady acceleration of 3m/s^2. How much velocity does it gain 10 seconds?
cluponka [151]
Acceleration measures how fast the speed changes over time.

So, over 10s, the aircraft's speed changes by 3\frac{m}{ s^{2} } * 10s

So, the change is 30\frac{m}{s}

Note that the one of the second units cancels out in the answer.
6 0
1 year ago
Read 2 more answers
How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236
Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between  non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

 

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(c) 7.03 = 3

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0
2 years ago
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
2 years ago
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