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3 years ago

6

Anyone wanna chitchat? i'm just bored

1 answer:

Mashcka [7]3 years ago

5 0

Answer:

yea me too you know your question is gunna get deleted bc it is not school related but I'm not gunna report it or anything

Explanation:

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If a diffraction grating has 3700 lines per cm, what is the spacing d between lines

Sonja [21]

So first we find the gap between the slits by the formula d=1/N

<span>N is number of lines per metre so 3700 line/cm = 370000 lines/m </span>
<span>So d=2.7*10^-6 </span>

<span>Now we use the formula dsin(angle)=n(wavelength) </span>

<span>d is the same </span>
<span>n is the order of the diffraction pattern </span>

<span>so wavelenth=dsin(angle)/n </span>
<span>=[(2.7*10^-6)*sin30]/3 </span>
<span>=4.5*10^-7 m</span>

7 0

3 years ago

sveticcg [70]

Answer:

I believe the answer is A and D.

I am unsure of C.

4 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium

ss7ja [257]

Answer:

The number of atoms are $1.86\times10^{8}$ .

Explanation:

Given that,

Diameter $D = 1.40\times10^{2}\ pm$

$D=1.40\times10^{2}\times10^{-12}\ m$

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

$Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}$

$Number\of\atoms =1.86\times10^{8}$

Hence, The number of atoms are $1.86\times10^{8}$ .

8 0

2 years ago

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T

NeX [460]

Answer

i dont know

Explanation:

4 0

2 years ago