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3 years ago

What happens to the diffraction pattern when the number of lines per centimeter of a diffraction grating is increased?

1 answer:

3 0

The separation between two neighboring vital maxima increments. As the quantity lines per centimeter expand, the detachment d between neighboring openings ends up noticeably littler. The point that determines primary maxima of a diffraction grinding is given by transgression = m/d, where is the wavelength and m = 0, 1, 2, 3. For settled estimations of m and the wavelength, this connection demonstrates that as d abatements, increments. For a settled screen area, this implies the separation between two neighboring chief maxima additionally increments.

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3 years ago

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Suppose the total momentum of two masses before a collision is 100 kg m/s. What is the total momentum of the two masses after th

soldier1979 [14.2K]

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3 years ago

A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl

Genrish500 [490]

Answer:

Explanation:

Given

mass of block $m=5.7\ kg$

at $t=0 s$

displacement is $x=-0.7\ m$

velocity $v=-0.8\ m/s$

acceleration $a=2.7\ m/s^2$

suppose $x=A\cos (\omega t+\phi )$ is the general equation of SHM

where A=amplitude

$\omega$ =natural frequency of oscillation

therefore velocity and acceleration is given by

$v=-A\omega \sin (\omega t+\phi )$

$a=A\omega ^2\cos (\omega t+\phi )$

for t=0

$-0.7=A\cos (\phi )---1$

$v=-0.8=-A\omega \sin(\phi)---2$

$a=2.7=-A\omega ^2\cos(\phi )----3$

divide 1 and 3 we get

$\omega ^2=\frac{27}{7}$

$\omega =\sqrt{\frac{27}{7}}$

Now square and 1 and 2 we get

$(0.7)^2+(\frac{0.8}{\omega })^2=A^2$

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3 years ago

The base ( foundation ) of building is made wider. Why ?

attashe74 [19]

Answer: It is increased to exert the pressure off the bywalls and on large area (brainliest pls)

Explanation: ...

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2 years ago

Kyle, a 95.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.

babymother [125]

Answer:

The initial speed of the ball was 26.2 m/s

Explanation:

When the football player is in the air at his maximum height the vertical component of velocity is zero, To obtain the horizontal velocity when the player catches the ball we need to apply the linear momentum conservation theorem:

$m_1*v_{o1}+m_2*v_{o2}=m_t*v_f\\0.430kg*(v)+m_2*(0)=mt*vt$

we need to obtain the time taken to go down.

$y=Y_o+v_o*t-\frac{1}{2}g*t^2\\\\0=0.589-4.9t^2\\solving:\\t_1=0.346s\\$

We have a horizontal displacement and the time taken to stop, so:

$v_f=\frac{d}{t}=\frac{0.0409m}{0.346s}=0.118m/s$

so:

$0.430kg*(v)+m_2*(0)=(m1+m2)*vt\\v=\frac{(95.0kg+0.430kg)*0.118m/s}{0.430kg}\\\\v=26.2m/s$

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3 years ago