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Bas_tet [7]
3 years ago
11

Light of wavelength 623.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 76.5 cm from the slit.

The distance on the screen between the second order minimum and the central maximum is 1.11 cm . What is the width a of the slit in micrometers (μm)?
Physics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

Explanation:

wave length of light λ = 623 x 10⁻⁹ m .

Distance of screen D = 76.5 x 10⁻² m

width of slit        =      d

Distance on the screen between the second order minimum and the central maximum       =  2  λ D / d

1.11 x 10⁻²  = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d

d =  ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²

= 85872.97 x  10⁻⁹

=  85.87297 x  10⁻⁶

= 85.87 μm

width a of the slit is = 85.87 μm

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What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant vel
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Answer:

b. mg ( μ · cos θ + sin θ)

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg ·  cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ      (N = wy = mg ·  cos θ)

F = μ · mg ·  cos θ  + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.

8 0
3 years ago
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