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11111nata11111 [884]
3 years ago
6

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las

er (λ = 633 nm) and a 0.12-mm-diameter pinhole. How far behind the pinhole should you place the viewing screen?
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

w=\frac{2.44 \lambda L}{D}

where \lambda is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m

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A car travels along a straight road to the east for
Korvikt [17]

Answer: 90 m/s, 70 m/s

Explanation:

Given

A car travels 100 m  east for 4 seconds and then turn around

goes back west for 50m  in 1 sec

Distance traveled in the east is 100\times 4=400\ m

Distance traveled in the west is 50\times 1=50\ m

Total distance=400+50=450\ m

total displacement=400-50=350\ m

total time t=4+1=5\ s

Average speed

\Rightarrow s_{avg}=\dfrac{\text{distance}}{\text{time}}=\dfrac{450}{5}\\\\\Rightarrow s_{avg}=90\ m/s

Average velocity

\Rightarrow v_{avg}=\dfrac{\text{displacement}}{\text{time}}=\dfrac{350}{5}\\\\\Rightarrow v_{avg}=70\ m/s

6 0
3 years ago
How far is a spring extended if it has 1.0 j of potential energy and its spring constant is 1,000 n/m?
DerKrebs [107]
E=F*d/2 = k*d * d/2 =>
d^2= 2*E/k

d= sqrt(2*E/k)=sqrt(2*1J/1000N/m)=sqrt(20m^2)/100=0.045 m = 45 mm
3 0
4 years ago
Which process causes the transfer of energy by air currents within the Earth's atmosphere?
ruslelena [56]

Answer:

Conduction, radiation and convection all play a role in moving heat between Earth's surface and the atmosphere. Since air is a poor conductor, most energy transfer by conduction occurs right near Earth's surface

8 0
3 years ago
In a nuclear experiment a proton with kinetic energy 3.0 MeV moves in a circular path in a uniformmagnetic field. What energy mu
dlinn [17]

Answer:

a)     K = 3 MeV   b)   K=  1.5 MeV

Explanation:

We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.

        F = q v x B

We can also write this equation based on the modules of the vectors

        F = qv B sin θ

With Newton's second law

       F = ma

       F = m v² / r

       q v B = m v² / r

       v = q B r / m

The kinetic energy is

       K = ½ m v²

Substituting

       K = ½ m (q B r/ m)²

       K = ½ B² r²  q² / m

       K = (½ B² R²)  q²/m

The amount in brackets does not change during the experiment

      K = A  q² / m

For the proton

     K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J

With this data we can find the amount we call A

    A = K  m/q²

    A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²

    A = 3.13 10⁻²

With this value we can write the equation

    K = 3.13 10⁻²  q² / m

Alpha particle

    m = 4 uma = 4 1.66 10⁻²⁷ kg

   K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷

   K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV

   K = 3 MeV

Deuteron

   K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷

   K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)

   K = 1.5 10⁶ eV

   K=  1.5 MeV

6 0
3 years ago
A box contains about 5.54 1021 hydrogen
Aleks [24]

Answer:E=33.71\ J

Explanation:

Given

No of atoms of hydrogen N=5.54\times 10^{21}

Temperature of room T=21^{\circ}\approx 294\ K

Thermal energy of the atoms is given by

E=\frac{3}{2}NkT

where k=boltzmann constant

E=\frac{3}{2}\times 5.54\times 10^{21}\times 1.38\times 10^{-23}\times 294

E=33.71\ J

Hence the energy of the atoms is 33.71\ J

5 0
3 years ago
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