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Gemiola [76]
3 years ago
5

A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that

the first completely dark fringes occur on the screen at distances of ±8.24 mm from the central bright fringe, and that the central bright fringe has an intensity of 2.00 W/m2 at its center. (a) How wide is the slit?
Physics
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

a = 2.275 10⁻⁴ m

Explanation:

This is a diffraction problem that is described by the equation

         a sin θ = m λ

         

The first dark minimum occurs for m = 1

         a = λ  / sin θ

The angle can be found by trigonometry,

       tan θ = y / x

       θ = tan⁻¹ y / x

Let's reduce the magnitudes to the SI system

        y = 8.24 mm = 8.24 10⁻³ m

        λ  = 625 nm = 625 10⁻⁹ m

       θ = tan⁻¹ 8.24 10⁻³ / 3.00

       θ = 0.002747 rad

We calculate

       a = 625 10⁻⁹ / sin 0.002747

       a = 2.275 10⁻⁴ m

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Scientists often use models to study the movement of continents. Why might scientists use a model to show this movement? A. Extr
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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
When friction slows a sliding block___
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Answer:

A

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When friction slows a sliding block, <u>the kinetic energy of the block is transformed into internal energy .</u>

<em>The frictional movement of two surfaces over one another leads to the conversion of some of their kinetic energies to another energy - heat or thermal energy. Hence, the temperatures of the objects are raised in the process. </em>

<u>Therefore, when a sliding block is slowed down due to friction, some of the kinetic energy of the block would be transformed into internal energy in the form of heat.</u>

The correct option is A.

6 0
3 years ago
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