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Korvikt [17]
2 years ago
12

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

wide-that is, the two first-order diffraction minima are separated by 1.40cm
What is the distance between the two second-order minima?
Physics
1 answer:
andrew11 [14]2 years ago
8 0

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

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vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

Ke=\frac{1}{2}Mass*Velocity^2

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules

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3 years ago
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At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

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s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

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v₀² + 2gh = v²

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