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GenaCL600 [577]
3 years ago
6

What is the approximate wavelength of a light whose first-order bright band forms a diffraction angle of 45.0° when it passes th

rough a diffraction grating that has 500.0 lines per mm?
236 nm
353 nm
943 nm
1414 nm
Physics
2 answers:
max2010maxim [7]3 years ago
7 0

Answer:

its D

Explanation:

edg2020

Bond [772]3 years ago
6 0

Answer:

D) 1414 nm

Explanation:

This is correct on Edge.

Using the equation dsin(angle)=n(wavelength), we can solve for wavelength.

First we must convert the 500 lines per mm to nm. We do this by 1/500, giving you 0.002. Then move the decimal over six places to the right, resulting in 2000.

Then by plugging in the other values, we have 2000sin(45)=1(wavelength).

N is one because we are just solving for a first-order band.

So 2000sin(45)=wavelength

By using a calculator, we can see that the wavelength equals approximately 1414nm.

I hope this helped. If it did, I would really appreciate a Brainliest!!

Have a great day:)

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Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f
Lunna [17]

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = \frac{1}{2}mv^{2}

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

mgh = \frac{1}{2}mv^{2}

final velocity of the child , v = \sqrt{2gh}

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

8 0
3 years ago
What are the dimensions of a frequency in physics
Burka [1]

Answer:

This is the information I can provide. I hope it helps

Explanation:

Frequency is measured in units of hertz (Hz) which is equal to one occurrence of a repeating event per second. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency.

7 0
4 years ago
Read 2 more answers
A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass of the man-bo
timama [110]

Answer:

x=1.25m

Explanation:

The <em>Center of mass </em>of the system is defined as the point where whole mass of the body is appeared to be  concentrated.

The center of mass of the system is given by

         x=  \frac{m1x1+m2x2}{m1+m2}      

where m1 is mass of man =60 kg

           m2 mass of board =20 kg

let the man be at the origin  x1 =0 , x2 =5m

by substituting in above formula

x =\frac{(60*0)+(20*5)}{60+20} = \frac{100}{80} =1.25 m

x=1.25m

So the center of mass of the system is at 1.25 m from man.

6 0
4 years ago
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An object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height abo
ki77a [65]

Answer:

11.09 m/s

Explanation:

Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.

The parameters given are:

Initial velocity U = ?

Final velocity V = 9.6 m/s

Acceleration due to gravity g = 9.8m/s^2

Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0

Using third equation of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8H

U^2 = 19.6H ........ (1)

Using the formula again for one fourth of its maximum height

9.6^2 = U^2 - 2 × 9.8 × H/4

92.16 = U^2 - 19.6/4H

92.16 = U^2 - 4.9H

U^2 = 92.16 + 4.9H ...... (2)

Substitute U^2 in equation (1) into equation (2)

19.6H = 92.16 + 4.9H

Collect the like terms

19.6H - 4.9H = 92.16

14.7H = 92.16

H = 92.16/14.7

H = 6.269

Substitute H into equation 2

U^2 = 92.16 + 4.9( 6.269)

U^2 = 92.16 + 30.72

U^2 = 122.88

U = 11.09 m/s

Therefore, the initial velocity of the object is 11.09 m/s

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