Is the change in velocity divided by the time needed for the change to occur.

A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?

Natali [406]

Three complete orders on each side of the m=0 order can be produced in addition to the m=0 order.

The ruling separation is d=1/(470mm-1)

$= 2.1 \times 10 {}^{ - 3}mm$

Diffraction lines occurs at an angle θ such that dsin=mλ,when λ is the wavelength and m is an integer.

Notice that for a given order,the line associated with a long wavelength is produced at a greater angle than the line associated with shorter wavelength.

we take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is,find the greater integer value of m for which mλ<d.

since,d/λ

$= 538 \times 10 {}^{ - 9} m/2.1 \times 10 {}^{ - 6}$

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

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4 0

2 years ago

Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th

jonny [76]

From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F = 83.4 × 10−11 /62500

F= 1.33 * 10^-14 N

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4 0

2 years ago

Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec

yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{3.74\times 9.8}{0.0161}$

k = 2276.52 N/m

The frequency of vibration is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}$

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0

3 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

Question 8 of 25

Dima020 [189]

Answer:

B.

Explanation:

8 0

3 years ago