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Anton [14]
3 years ago
10

Using a 685 nm wavelength laser, you form the diffraction pattern of a 1.11 mm wide slit on a screen. You measure on the screen

that the 11th dark fringe is 9.85 cm away from the center of the central maximum. How far is the screen located from the slit
Physics
1 answer:
velikii [3]3 years ago
5 0

Answer:

13.8 m

Explanation:

Dark fringes are formed in a single slit experiment due to destructive interference that occurs due to interference.

The position of these dark fringes formed on a screen is given by:

y = \frac{\lambda }{d} (m + 1/2)D

where y = position of  mth minimum

m = order of the minimum

D = distance of the slit from the screen

d = width of the slit

λ = wavelength of the light used

We need to find D:

D = \frac{yd} {\lambda (m + 1/2)} \\

From the question:

m = 11

y = 9.85 cm = 0.0985 m

λ  = 6.83 * 10^{-7} m

d = 1.11 mm = 0.0011 m

Therefore:

D = \frac{0.0985 *0.0011} {6.83 * 10^{-7} *(11 + 1/2)} \\\\D = \frac{0.00010835} {6.83 * 10^{-7}  * (23/2)} \\\\D = 13.8 m

The slit is 13.8 m far from the screen

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What conclusions can you draw about the connection between the Sun’s altitude and the latitude of the observer on seasonal chang
Cerrena [4.2K]

Answer:

Explanation:

Altitude of the Sun and the latitude position on the earth play an important role in the season change on the earth.

When the altitude of the sun is high then the average temperature of the earth is higher because the luminous intensity of the sun rays is higher due to the focusing of high energy sun rays over a small area.

But when the sun is at higher altitudes we receive less denser rays of the sun and hence we have less heat on the earth on an average.

  • But despite of the altitude some places on the earth have distinct temperature than the other place at the same time of the year. This is due to their latitudinal location. The places near the equator are warmer most of the times throughout the year because they receive the most direct rays while the poles receive slanting rays and hence are colder even in summer when the earth is at lower altitudes.
6 0
3 years ago
A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They
denis-greek [22]

Answer:

  v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

         Em₀ = U = m g h

Final point. Lower

        Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

         v² = 2gh

         v = √ 2gh

Let's calculate

         v = √ (2 9.8 150)

         v = 54.2 m / s

3 0
3 years ago
A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.
ycow [4]

Answer:

σ = 3.402 KPa ,  γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

Find the shearing stress, shearing strain and  shear modulus.​

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

- The shear effect results in a stress in the gelatin block.

- The formulation of stress ( σ ) is given below:

                        σ = F / A

Where,

           A : The surface area of the object that experiences the shear force.

- The top surface have the following dimensions:

          A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

                     σ = 49 / 0.0144

                     σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

                    γ = x / L

                    γ = 10 / 40

                    γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- It is mathematically expressed as a ratio of shear stress  ( σ ) and shear strain ( γ ):

                   G =  σ / γ

                   G = 3.402 / 0.25

                   G = 13.608 KPa

7 0
3 years ago
A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be
vodomira [7]

Answer:

Explanation:

i )

When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2

So charged stored in it will remain unchanged .

ii )

Potential difference = charge / capacitance

in the first case potential difference = Q / C

in the second case potential difference = Q / 2C

So potential difference becomes half .

iii ) electric field = potential diff / plate separation

in the first case electric field = Q / (d x C )

in the second case electric field = 2 Q / (d x 2C)

= Q / (d  x C )

So electric field remains unchanged .

iv)

energy stored in first case = Q² / 2C

In the second case energy stored = Q² / 2x2C

so energy stored becomes half .

4 0
3 years ago
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