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mylen [45]
3 years ago
10

In a Young’s double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe i

s 5.4. Find the ratio d/ of the slit separation d to the wavelength of the light. (d / λ) =?
Physics
1 answer:
AVprozaik [17]3 years ago
4 0

Answer:

d / λ = 26.7

Explanation:

In Young's double slit experiment, constructive interference is described by the expression

    d sin θ = m λ

In the case of destructive interference we must add half wavelength (λ/2)

    d siyn θ = (m + ½) λ

Let's clear

     d / λ = (m + ½) / sin θ

Let's calculate

    d / λ = (2+ ½) / sin 5.4

    d / λ = 5 / (2 sin 5.4)

    d / λ = 26.7

You might be interested in
Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0
2 years ago
8. While taking a measurement, Ajay put the 2nd mark of the scale to the edge of the line and the mark that pointed to the end o
Leni [432]

Answer:

The length of line is 78 cm or 0.78 m.

Explanation:

initial reading 2 mark

final reading 80 cm

The length of the line

= final reading - initial reading

= 80 - 2

= 78 cm

1 cm = 0.01  m

So, 78 cm = 0.78 m

4 0
2 years ago
A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s
AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight W = 150 N

Length of rope r = 4.1 m

Initial speed of ball v = 5.5 \frac{m}{s}

For finding the tension in the rope,

First find the mass of rod,

mg = 150                          ( g = 9.8 \frac{m}{s^{2} } )

  m = \frac{150}{9.8}

  m = 15.3 kg

Tension in the rope is,

  T = mg + \frac{mv^{2} }{r}

  T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}

  T = 262.88 N

Therefore, the tension in the rope is 262.88 N

7 0
3 years ago
O
Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0
3 years ago
Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop
Sonja [21]
For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
8 0
3 years ago
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