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A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha

Ber [7]

Answer:

The light used has a wavelenght of 4.51×10^-7 m.

Explanation:

let:

n be the order fringe

Ф be the angle that the light makes

d is the slit spacing of the grating

λ be the wavelength of the light

then, by Bragg's law:

n×λ = d×sin(Ф)

λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

= 4.51×10^-5 cm

≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

7 0

3 years ago

Convert 2 days into second

eduard

Answer:

172,000 second 

I'M NOT SURE THAT THIS IS RIGHT OR WRONG IF IT'S WRONG THEN SORRY 

3 0

2 years ago

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In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re

Irina-Kira [14]

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

1/2 m r² x α = torque

As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .

6 0

2 years ago

A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

8 0

3 years ago

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The Kapandu boys dormitory in the UPNG is going up in flames. Rachel, a former student of Physics, who

Tom [10]

Answer:

1108.464 N of force

Explanation:

diameter of water hose = 70 cm = 0.7 m

radius = 0.7/2 = 0.35 m

volumetric flow rate Q = 420 L/min

1 L = 0.001 m^3

1 min = 60 s

therefore,

Q = 420 L/min = (420 x 0.001)/60 = 0.007 m^3/s

Area A of fire hose = π $r^{2}$ = 3.142 x $0.35^{2}$ = 0.38 m^2

From continuity equation, Q = AV

where V1 is the velocity of the water through the pipe, and A1 is the area of the pipe.

Q = A1V1

0.007 = 0.38V1

V1 = 0.007/0.38 = 0.018 m/s.

Nozzle diameter = 0.75 cm = 0.0075 m

radius = 0.00375

Area = π $r^{2}$ = 3.142 x $0.00375^{2}$ = 4.42 x $10^{-5}$ m^2

velocity of water through the nozzle will be

V2 = Q/A2 = 0.007 ÷ (4.42 x $10^{-5}$ ) = 158.37 m/s

From

F = ρQ(v2 - v1)

Where,

F = force exerted

p = density of water = 1000 kg/m^3

F = 1000 x 0.007 x (158.37 - 0.018) = 1108.464 N of force

4 0

3 years ago

ANSWER IN LESS THAN A MINUTE!! EASYY!​

ANSWER IN LESS THAN A MINUTE!! EASYY!