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krok68 [10]
3 years ago
14

A diffraction grating with 270 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. 1. A

t what angle from the beam axis will the fourth order peak occur if the tube emits light with wavelength of 665.0 nm?2. At what angle will the second order peak occur?
Physics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

1) θ = 45.91°

2) θ = 21.04°

Explanation:

We are given;

Wavelength; λ = 665 nm = 6.65 × 10^(-7) m.

Distance between slits; d = 1mm/270 = 1/270 mm = (1/270) × 10^(-3) m

1) To find the angle, we will use the formula;

d sin θ = mλ

Where m is the order of peak which in this question is 4.

Thus, we have;

sin θ = mλ/d

sin θ = (4 × 6.65 × 10^(-7))/((1/270) × 10^(-3))

sin θ = 0.7183

θ = sin^(-1) 0.7183

θ = 45.91°

2) Similarly, d sin θ = mλ

Where m is the order of peak which in this question is 2. Thus;

sin θ = (2 × 6.65 × 10^(-7))/((1/270) × 10^(-3))

sin θ = 0.3591

θ = sin^(-1) 0.3591

θ = 21.04°

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If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.

Explanation:

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What simple machines combine to make a shovel which is a compound
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A truck drives with a constant linear speed v_iv i ​ v, start subscript, i, end subscript down a road with two curves. The first
Alex

Answer:

(C) Decreases by factor of 3

Explanation:

Centripetal acceleration is given by

a = \dfrac{v^2}{r}

where <em>v</em> is the linear velocity and <em>r</em> is the radius of the curve.

Let the centripetal acceleration on the curve of radius <em>R</em> be a_1.

Then

a_1 = \dfrac{v_i^2}{R}

Let the centripetal acceleration on the curve of radius 3<em>R</em> be a_2.

Then

a_2 = \dfrac{v_i^2}{3R} = \dfrac{1}{3}\dfrac{v_i^2}{R} = \dfrac{1}{3}a_1

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7 0
3 years ago
Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele
nekit [7.7K]

Answer:

0.536\sqrt{\frac{GM}{R}}

Explanation:

We are given that

Mass of one  asteroid 1,m_1=M

Mass of asteroid 2,m_2=1.97 M

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

u=0

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

(m_1+m_2)u=m_1v_1+m_2v_2

(M+1.97 M)\times 0=Mv_1+1.97Mv_2

Mv_1=-1.97 Mv_2

v_1=-1.97v_2

According to law of conservation of energy

Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2

1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)

1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2

v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}

v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}

v_2=0.536\sqrt{\frac{GM}{R}}

Hence, the speed of second asteroid =0.536\sqrt{\frac{GM}{R}}

8 0
3 years ago
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