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olasank [31]
3 years ago
5

Consider a diffraction grating of width 5cm with slits of width 0.0001 cm separated by distance of 0.0002cm 1:what is the corres

ponding grating element 2:how many orders will be observed at a wavelength of 0.000055 cm 3:calculate the dispersion in the different orders
Physics
1 answer:
sashaice [31]3 years ago
3 0

Answer:

a) d= 0.0002cm

b) m=3

Explanation:

From the question we are told that:

Width of diffraction grating W_g= 5cm

Width of silt w_s 0.0001 cm

Distance d= 0.0002cm

a)

Generally the grafting element is the distance b/w  the silts

d= 0.0002cm

Therefore the grafting element is

d= 0.0002cm

b)

Generally the equation for diffraction state is mathematically given by

 dsin\theta=m\lambda

Since

 sin\theta \leq 1

Therefore

 m \leq \frac{d}{\lambda}

 m \leq \frac{0.0002}{0.000055}

 m \leq 3.6

Therefore orders observed with \lambda is

 m=3

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joja [24]

Answer: Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.

Explanation: In the 1600s, Boyle measured the volumes of gases at different pressures. Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.

3 0
3 years ago
when do you use cos and sin in situations like these? is horizontal always cos and vertical always sin?
Andreas93 [3]

Answer:

yes

Explanation:

this is simple

the horizontal line is adjacent

the vertical line is opposite

recall that cos x=adj/hyp

adj=hyp(cos x)

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8 0
3 years ago
A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio
goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

3 0
3 years ago
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

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I_{cm}=MR^2

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v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

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3 years ago
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