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iragen [17]
4 years ago
14

Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

n is observed on a screen located at a distance of 3.0 m from the slits. Where will the second order dark fringe above the center of central bright fringe will form?
Physics
1 answer:
goblinko [34]4 years ago
7 0

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d  

Putting the  values given above

distance = \frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}

= 134.77 x 10⁻³ m

= 134.77 mm.

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A circular ride has a radius of 32 m If the time of one revolution of a rider is 0.98 s what is the speed of the rider?
11111nata11111 [884]

Given that,

The radius of a circular path, r = 32 m

The time of one revolution of a rider is 0.98 s.

To find,

The speed of the rider.

Solution,

Let v is the speed of the rider. Speed is equal to total distance divided by time taken.

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 32}{0.98}\\\\v=205.16\ m/s

So, the speed of the rider is 205.16 m/s.

6 0
3 years ago
The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant
KIM [24]

Answer: 50 km, 0, 27.78 m/s

Explanation:

Given

Circumference of the track is 12.5\ km

Speed of car is 100\ km/h

Car drives for 30\ \text{minute}\ or\ 0.5\ hr

(a)Distance traveled is

\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km

(b)displacement of the car

It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.

Therefore, the displacement of the car is zero.

(c)To convert kmph to m/s, multiply the entity by \frac{5}{18}

\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s

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3 years ago
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3 years ago
1. Summarize What are the four conditions of a<br>purely competitive market?​
bearhunter [10]

Answer:

Explanation:

1. Many buyers and sellers participate in the market.

2. Sellers offer identical products.

3. Buyers and sellers are well informed about products.

4. Sellers are able to enter and exit the market freely.

5 0
3 years ago
A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the
Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = 9.40\times 10^{4}\ Pa

Pressure in the smaller pipe, P' = 2.80\times 10^{4}\ Pa

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'

18^{2}v = 9^v'

v' = 4v

Now,

By using Bernoulli's eqn:

P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'

where

h = h'

\rho = 10^{3}\ kg/m^{3}

9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}

6.6\times 10^{4}  = \frac{1}{2}\rho 15v^{2}

v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s

Now, the rate of flow is given by:

\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av

\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s

3 0
3 years ago
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