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lina2011 [118]
3 years ago
12

Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da

rk fringe appears 3 mm from the central maximum. What is the width of the slit?
Physics
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light \lambda =550nm=550\times 10^{-9}m

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So x=3\times 10^{-3}m

We have to find the width of the slit

For the first order wavelength is equal to \lambda =\frac{x}{D}\times a, here a width of slit

So a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm

So width of the slit will be equal to 1.47 mm

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The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
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The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

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Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such
Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity  = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0
3 years ago
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