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mariarad [96]
3 years ago
7

A Young's double-slit experiment is performed using light that has a wavelength of 631 nm. The separation between the slits is 5

.25E-5 m. Calculate the angle that locates the first-order bright fringes on the screen.
Physics
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer:

 θ = 1.21  10⁻² rad

Explanation:

The double slit experiment is described for constructive interference

           d sin θ = m λ

where d is the separation between the slits, λ the wavelength used, m a stereo that represents the order of interference

The first order of interference corresponds to m = 1 and the separation of the slits is d = 5.2 10⁻⁵ m

        sin θ = m λ / d

        θ = sin⁻¹ (m λ / d)

     

let's calculate

          θ = sin⁻¹ (1 631 10⁻⁹ / 5.2 10⁻⁵)

          θ = 1.21  10⁻² rad

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According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
Bond [772]

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
3 years ago
What do we call fixed point around which a lever pivots?
xxTIMURxx [149]
Answer is A) Fulcrum

The fixed point that a lever rotates around is called the fulcrum.
4 0
2 years ago
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A crowbar having the length of 1.75m is used to balance a load between the fulcrum and the load is 0.5m calculate MA
aliina [53]

Answer:

1.70

Explanation:

substract 1.75-0.5

3 0
3 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
3 years ago
a parking lot is going to be 60m wide and 240m long which dimensions could be used please please help
Bezzdna [24]

Answer: The area of the parking lot is 14,400 meters squared.

Explanation:

We have the dimensions of the parking lot.

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The units used here are meters.

Now, if we want to know the area of the parking lot is equal to the product between the length and the width:

A = 60m*240m = 14,400 m^2

The area of the parking lot is 14,400 meters squared.

3 0
4 years ago
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