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navik [9.2K]
4 years ago
9

Light from an argon laser strikes a diffraction grating that has 7,735 grooves per centimeter. The central and first-order princ

ipal maxima are separated by 0.488 m on a wall 1.73 m from the grating. Determine the wavelength of the laser light.
Physics
1 answer:
Leto [7]4 years ago
4 0

Answer:

wavelength of the laser light is 351nm

Explanation:

The split seperation is

d = 0.01 / 7735

= 1293nm

The angle of the first order maxima

θ = tan⁻¹ (0.488/1.73)

= 15.75°

we have,

sin \theta = \frac{m \lambda}{d} \\\lambda = \frac{d sin \theta}{m} \\\lambda = \frac{1293 \times sin15.75}{1} \\= 350.97nm

≅ 351nm

wavelength of the laser light is 351nm

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BigorU [14]
The formula for orbital speed v is    v=(G*Me/r)^1/2
Wher G= 6.67E-11,    Me= 6E24,    r= Re+h= 6.4E6+740000
putting values in the formula we get

v= 7486.7 m/s     or  v= 7.4867 km/s
3 0
4 years ago
When the diaphragm contracts , air pressure in the chest increases . True or False ?
nexus9112 [7]

Answer: False

 

When the diaphragm contracts, the muscles will also contract and pull upward and increase the size of the thoracic cavity thus decreases air pressure inside during inspiration. After the diaphragm contracts, it goes to relaxation, the muscles will also relaxed. It gets looser and return to its original position higher up in the chest. This increase the pressure in the chest, which force the air in the lungs out through the nose.

 

 

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3 years ago
Read 2 more answers
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
4 years ago
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Harlamova29_29 [7]

Answer:

7506 cal

Explanation:

Given: Mass of water = m = 13.9 grams.

Heat energy required to determine the conversion of 13.9 grams of water to steam or vapor is m L,

where L = Latent heat of vaporization of water=540 cal/g.

Q = mL = (13.9)(540) = 7,506 cal.

6 0
4 years ago
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Aleksandr-060686 [28]

Answer:

Light travels very fast

Explanation:

Speed of light is normally estimated as 299704644.54 m/s in air hence radar systems have always been used to measure the distance travelled per unit time. By Galileo experimenting to get the speed of light in the above mentioned set-up, it would be difficult to estimate the speed of light. Therefore, it was obvious that Galileo could not estimate the speed of light. This is because the speed of light is very high. In vacuum, the speed of light is estimated as 299,792,458 m/s and this can't be accurately measured in such mentioned set up by Galileo.

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