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Elis [28]
3 years ago
9

Light from a helium-neon laser (λ = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a sc

reen 3.00 m behind the slits. Eleven bright fringes are seen, spanning a distance of 52.0 mm. What is the spacing (in mm) between the slits?
Physics
1 answer:
stepladder [879]3 years ago
7 0

Answer:

The spacing between the slits is <u>0.40233 mm</u>

Explanation:

Let 'd' be the distance between the slits and 'D' be the distance of the screen from the plane of slits.

We know that , in an interference pattern the fringe(dark or bright) width is given by -

f = \dfrac{Dλ}{d}

For 11 bright fringes , if its total spanning distance is 52.0 mm , then for one bright fringe , the fringe width(f) would be -

f = \dfrac{52}{11} = 4.72 mm

∴ From above equations ,

\dfrac{Dλ}{d} = 4.72 mm = (4.72 × 10^{-3}[/teλx]) mHere, D = 3 m           λ = 633 nm =(633 × [tex]10^{-9}) m

Substituting the values in above equation -

\dfrac{3×633×10^{-9} }{d} = 4.72 ×10^{-3}

∴ d = 0.0004023 m =0.4023 mm

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A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the
Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs  with respect to the axis of the rotation is expressed by the formula:

L_i ^ { ^  \to } = mp ( r_y } ^ { ^  \to }  * v_{pi}  ^ { ^  \to } )    ----- equation (1)

The position vector can now be :

x ^ { ^  \to } = - 0.4 \ j \ m

Also, given that :

v_{p,i}  ^ { ^  \to } = (280 \ i - 350 \  j)  \ m/s

Replacing the value into above equation (1); we have:

L_i ^ { ^  \to } =0.012 ((- \ 0.4 \  j) *(280 \ i - 350 \ j ))

L_i ^ { ^  \to } =0.012 * 112 \  k    (by using cross product )

L_i ^ { ^  \to } = 1.344 k` \  \  kg  m^2 s^{-1}

However; the moment of inertia of the rod about the axis of rotation is :

I_{rod} = \frac{1}{3}m_rl^2  \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \  kg  \ m^2

Also, the moment of inertia of the pellet about the axis of rotation is:

I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2  \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2

So, the moment of inertia of the rod +pellet system is:

I = I_{rod}+I_{pellet}

I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2

I = 2.6686 \  kg. m^2

The final angular momentum is :

L_f ^ {^ \to}  = I \omega { ^ {^ \to} } = 2.6686 \  \omega  ^ {^ \to}

The angular velocity of the rod \omega is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

L_f ^ {^ \to} = L_i ^ { ^ \to}

2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}

\omega  ^ {^ \to} = \frac{1.344 \ k`}{2.6686}

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

7 0
3 years ago
Marisa’s car accelerates at an average rate of 2.6m/s^2. Calculate how long it takes her car to accelerate from 24.6m/s to 26.8m
djverab [1.8K]

given info is... Acceleration(a)=2.6m/s^2

                       final velocity(v)=26.8m/s

                       initial velocity(u)=24.6m/s

need to find.... time(t)=?

a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\

t=\frac{v-u}{a}

t=\frac{26.8-24.6}{2.6}

t=0.846s

Explanation:

8 0
3 years ago
An object floats in water with 5 8 of its volume submerged. The ratio of the density of the object to that of water is
ki77a [65]

Answer:

\dfrac{5}{8}

Explanation:

m = Mass of object = \rho v

m' = Mass of water = \rho' v'

\rho = Density of object

\rho' = Density of water

Weight of the water displaced is the force in the case of floating objects

According to the question

v'=\dfrac{5}{8}v

In the case of floating objects

W=W'\\\Rightarrow mg=m'g\\\Rightarrow \rho vg=\rho'v'g\\\Rightarrow \rho v=\rho' \dfrac{5}{8}vg\\\Rightarrow \rho=\rho' \dfrac{5}{8}\\\Rightarrow \dfrac{\rho}{\rho'}=\dfrac{5}{8}

The ratio of the density of the object to that of water is \dfrac{5}{8}

3 0
3 years ago
What objects have orbital paths? A. the sun and stars
suter [353]

Answer:

The planets and moons.

Explanation:

Planets follow an elliptical path around the sun (kinda oval shaped). Moons do the same to planets.

7 0
2 years ago
Light waves from the sun can travel through outer space and reach the earth. This suggests that light is a(n)
garik1379 [7]

Answer:

D) Electromagnetic waves

Explanation:

This is because with  compression waves and longitudinal waves a medium is required such as air. however in a vacuum this will not work

8 0
3 years ago
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