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3 years ago

In Young’s double-slit experiment using monochromatic light, the interference pattern consists of a central _____.

Physics

2 answers:

svetoff [14.1K]3 years ago

8 0

Answer:

B)bright band with alternate dark and bright bands on either side

Explanation:

In young's double slit experiment we verify the wave characteristic of light

In this wave phenomenon we can see that on screen there is path difference of waves coming from two slits of light source due to which the total intensity of light on the screen will change accordingly

At the central point of the screen the path difference of light is ZERO

due to this the net intensity of the light on the central part of the screen will be maximum.

Since it is a monochromatic light experiment so now when we move on either side on the central point the path difference will change and hence we will get the maximum and minimum intensity alternatively on the screen

So correct answer will be

B)bright band with alternate dark and bright bands on either side

Pie3 years ago

5 0

In young's double slit experiment that uses monochromatic light the interference pattern formed has Central bright band with alternate dark and bright band That is option B

Explanation-

In this young experiment two small slits Namely a and b are formed on the screen and a monochromatic light is focused on them. Wavelets come out of of this lets scintillating and overlapping each other. This we get from huyginns principal, thus forming alternate dark and bright bands with bright band at center and all bands are about one meter apart.

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A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Olenka [21]

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground. So we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

$a=g=9.8 m/s^2$ is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

$v=0+(9.8)(0.75)=7.35 m/s$

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3 years ago

Which image illustrates refraction?

maks197457 [2]

Answer:

B illustrates refraction

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3 years ago

A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym

user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

L_(p,i) = m*V_pi*R

L_(p,i) = 3.6*3.3*0.44

L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

L_(c,i) = I*w_i

L_(c,i) = 0.5*M*R^2*0

L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

L_i = L_(p,i) + L_(c,i)

L_i = 5.2272 + 0

L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

L_f = L_(p,f) + L_(c,f)

L_f = I_p*w_f + I_c*w_f

L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

L_i = L_f

5.2272 = w_f*(I_p + I_c)

w_f = 5.2272/ R^2*(m + 0.5M)

Plug in values:

w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)

w_f = 5.2272/ 5.05296

w_f = 1.0345 rad/s

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3 years ago

Natalie lifts a 15-kg rock from the ground onto a 1.5 meter high wall. what is the amount of potential energy she has given the

Zina [86]

The amount of gravitational potential energy acquired by the rock is equal to:
$\Delta U = mg \Delta h$
where
m is the mass of the rock
g is the gravitational acceleration
$\Delta h$ is the increase in height of the rock

Substituting the data of the problem, we find
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So, Natalie gave 220.7 J of energy to the rock.

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3 years ago

Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain

Zepler [3.9K]

Answer: car 1 is going how fast

Explanation: no need to answer without speed I won't know distance.

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