1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
solmaris [256]
1 year ago
13

In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. what happens to the width

of the central bright fringe in the resulting diffraction pattern?
Physics
1 answer:
tamaranim1 [39]1 year ago
7 0

The width of the central bright fringe <u>becomes wider</u> in the resulting diffraction pattern of a single-slit diffraction experiment.

<h3>What is diffracted light?</h3>

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

<h3>What precisely is single slit diffractive?</h3>

The single-slit diffraction experiment allows us to examine the phenomena of light bending, or diffraction, which enables coherent light from a source to interfere with itself and generate the diffraction pattern, a recognizable pattern on the screen. When the sources are small enough to be relative to the wavelength of light, diffraction is seen.

Learn more about diffraction

brainly.com/question/8645206

#SPJ4

You might be interested in
Help asap physics homework, i'll give brainliest
Bess [88]

Answer:bippity boppity yee

Explanation:

5 0
3 years ago
Read 2 more answers
Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478
klasskru [66]

Answer:

a) 2.5 m/s²

b) 6.12 m/s

Explanation:

Tension of rope = T = 356N

Weight of material = W = 478 N

Distance from the ground = s = 7.5 m

Acceleration due to gravity = g = 9.81 m/s²

Mass of material = m = 478/9.81 = 48.72

Final velocity before the bundle hits the ground = v

Initial velocity = u = 0

Acceleration experienced by the material when being lowered = a

a) W-T = ma

⇒478-356 = 48.72×a

\Rightarrow \frac{122}{48.72} = a

⇒a = 2.5 m/s²

∴ Acceleration achieved by the material is 2.5 m/s²

b) v²-u² = 2as

⇒v²-0 = 2×2.5×7.5

⇒v² = 37.5

⇒v = 6.12 m/s

∴ Velocity of the material before hitting the ground is 6.12 m/s

5 0
3 years ago
A Tennis ball falls from a height 40m above the ground the ball rebounds
worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

4 0
3 years ago
If we have less power, we most likely have
katrin [286]
<span>the same amount of work being done over a longer period of time.</span>
3 0
3 years ago
Read 2 more answers
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
Other questions:
  • 3) A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.01
    8·1 answer
  • A physics student throws a ball straight up. The student catches the ball in exactly the same place from which it was released.
    15·1 answer
  • Refer to the following diagram to answer this question
    9·2 answers
  • An elevator weighing 2.00 x 10 5 N is supported by a steel cable. What is the tension in the cable when the elevator is accelera
    11·1 answer
  • A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s
    15·1 answer
  • PLEASEEEEE HELPPPPPP MEEEEEE I WILL MARK AS BRAINLIST PLEASEEEE
    11·1 answer
  • A car accelerates uniformly from 0 km/hr to 60 km/hr in 4.5 seconds. Which one of the following choices best represents the acce
    15·1 answer
  • The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what
    12·1 answer
  • Que es el periodo de un pendulo
    5·2 answers
  • What do you mean by potential energy of an object is 60J​
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!