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BartSMP [9]
3 years ago
13

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second

diffraction minimum and the central maximum is 1.40 cm. (a) Calculate the angle of diffraction θ of the second minimum.(b) Find the width of the slit.
Physics
1 answer:
LiRa [457]3 years ago
8 0

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

a sin\theta = m\lambda

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

\theta = Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

sin\theta = \frac{y}{d}

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

\theta = sin^{-1}(\frac{y}{d})

\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})

\theta = 0.2673\°

PART B) Equation both equations we have

a sin\theta = m\lambda

a \frac{y}{d} = m\lambda

Re-arrange to find a,

a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}

a = 1.65*10^{-4}m

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid.
kolbaska11 [484]

Answer:

The terminal velocity of the diver is 115 m/s = 414 km/hr

Explanation:

At terminal velocity,

Fnet = mg - Fd = 0

Drag force, Fd = cρAv²/2

mg = cρAv²/2

Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by

v = √(2mg/ρcA)

where m = mass of body falling through fluid = 80 kg

g = acceleration due to gravity = 9.8 m/s²

ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³

c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7

A = the area of the diver facing the fluid = 0.14 m²

v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr

5 0
3 years ago
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Explanation:

5 0
2 years ago
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According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the
slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

A=\dfrac{\pi}{4}\times d^2

Put the value into the formula

A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2

A=1.96\times10^{-9}\ m^2

We need to calculate the magnitude of the force

Using formula of force

F=\sigma A

Put the value into the formula

F=196\times10^{6}\times1.96\times10^{-9}

F=0.38416\ N

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

strain=\dfrac{\Delta l}{l_{0}}

\Delta l=strain\times l_{0}

Put the value into the formula

\Delta l=0.380\times l_{0}

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

l=l_{0}+\Delta l

Put the value into the formula

I=l_{0}+0.380\times l_{0}

l=1.38l_{0}

l_{0}=\dfrac{l}{1.38}

l_{0}=\dfrac{12\times10^{-2}}{1.38}

l_{0}=0.0869\ m

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

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bazaltina [42]
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
4 0
3 years ago
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