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crimeas [40]
2 years ago
5

When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. what is the

wavelength of the light?
Physics
2 answers:
zhannawk [14.2K]2 years ago
6 0

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

ans: 633nm

Sergio [31]2 years ago
3 0

Answer:

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

answer*= 633nm

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In two separate double slit experiments, an interference pattern is observed on a screen. In the first experiment, violet light
sashaice [31]

Answer:

\lambda_3 = 4.72*10^{-7} m

Explanation:

given data:

wavelength \lambda = 708nm = 708*10^{-9} m

using the following relation:

y = \frac{mL\lambda}{d}

according to the given information

second and third dark fringe is at same location. so

y_2 = y_3

\frac{m_2L\lambda_2}{d} = \frac{m_3L\lambda_3}{d}

m_2\lambda_2 = m_3\lambda_3

2*708*10^{-9} = 3*\lambda_3

\lambda_3 = \frac{2*708*10^{-9}}{3}

\lambda_3 = 4.72*10^{-7} m

4 0
2 years ago
A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 cm and a cross-sectional area of 15 cm2. T
stealth61 [152]
<span>A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?</span>
3 0
2 years ago
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Two objects, each of weight W, hang vertically by spring scales as shown in the figure. The pulleys and the strings attached to
valentinak56 [21]

The reading on the scale is the tension on the string that connects the two objecst. In order to support the blocks it must pull the weights by a force magnitude of W. So, the tension of the rope is W. Therefore, the reading on the scale is W, D.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0
2 years ago
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The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec
saveliy_v [14]

The static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta.

The given parameters;

  • <em>mass of the block, = M</em>
  • <em>coefficient of static friction in section 1, = </em>\mu_s_1<em />
  • <em>angle of inclination of the plane, = θ</em>

<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta

Thus, the static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta

Learn more here:brainly.com/question/17237604

3 0
2 years ago
James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m
Over [174]

Answer:

4.1\cdot 10^8 N

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

p=p_0 + \rho g h

where

p_0 =1.01\cdot 10^5 Pa is the atmospheric pressure

\rho = 1000 kg/m^3 is the water density

g=9.8 m/s^2 is the gravitational acceleration

h=11,000 m is the depth

Substituting,

p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2

And so, the inward force exerted on it is

F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N

8 0
2 years ago
Read 2 more answers
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