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FinnZ [79.3K]
3 years ago
13

Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5^degree away

from the central fringe. What is the separation between the double slits? a) 5.0 mu m b) 10 mu m c) 15 mu m d) 20 mu m e) 25 mu m
Physics
1 answer:
Anon25 [30]3 years ago
4 0

Answer:

The correct option is C

Explanation:

From the question we are told that

   The wavelength is  \lambda  =  575 *10^{-9} \  m

    The  angle is  \theta  =  6.5^o

    The order of maxima  is  n =  3

Generally for  constructive interference

       dsin \theta  =  n  *  \lambda

=>   d =  \frac{n  *  \lambda }{  sin \theta }

=>   d =  \frac{3   *  575 *10^{-9} }{  sin 6.5  }

=>   d = 15.24 *10^{-6} \  m

=>  d = 15 \mu  m

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Gravitational potential energy i think
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2 years ago
A helicopter ambulance flew from one hospital to another in a straight line. The pilot had to change speed several times due to
Olenka [21]

Answer:

Explanation:

All the rest of the information is extraneous. The only 2 things you have to know are

d = 20 km

t = 8 minutes = 8/60 hours = 0.13333333

So the speed is s = d/t

s = 20/0.1333333 = 150 km/hour

Note: you have not specified what units the speed is. I suppose you could answer 20/8 = 2.5 km/min

4 0
3 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
2 years ago
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?​
Xelga [282]
<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

<h3>And,</h3>

30 cm = 0.3 m (distance)

<h3>So,</h3>

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

\small \boxed{work \: done \:  by \: Ryan \:  = 3 \: Joules}

4 0
2 years ago
Can someone help with this???
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no BECQUSE POSUM BROOB SHSHSJ

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