What Is The frequency of an X Ray That had A Wavelength of 1.5* 10^-9

1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit

Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:

Given as

y = nDλ/w Eqn 1

where

w = width of slit

D = distance to screen

λ = wavelength of light

n = order number

Making x the subject of the formula gives,

w = nDλ/y

Given

y = 0.0149 m

D = 0.555 m

λ = 588 x 10-9 m

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1, y = nDλ/w

given, D = 27cm = 0.27m

λ = 632 x 10-9 m

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0

2 years ago

15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and di

lozanna [386]

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

B = μ₀ N/L I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

B = (μ₀ I / L) N

the amount between paracentesis constant, in the case of 4 loop the field is worth

B = cte 4

N B

4 4 cte

3 3 cte

2 2 cte

1 1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

6 0

3 years ago

Please help me .finish this paper

Evgesh-ka [11]

Solution-1:-

$\boxed{\sf \dfrac{10\times 1000}{60\times 60}}$

Solution:-2

$\boxed{\sf Sodium\:and\:Potassium}$

Solution:-3

$\boxed{\sf 320m}$

Solution:-4

$\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}$

Solution:-5

$\boxed{\sf Mg_3N_2}$

Solution:-6

$\boxed{\sf Grapes\:and\:Rambutan}$

Solution:-7

$\boxed{\sf {}^{}_{}N}$

Solution:-8

$\boxed{\sf Galactuse}$

Solution:-9

$\boxed{\sf Y-X}$

Solution:-10

$\boxed{\sf Cell\:wall}$

7 0

2 years ago

13. Determine the kinetic energy of a 2000g roller coaster car that is moving with a speed of 2m/s.

zubka84 [21]

Answer:

4 Joules (if you mean 2000 grams)

4000 Joules (if you mean 2000 kilograms)

Explanation:

<h2>PLEASE MARK AS BRAINLIEST!!!!!</h2>

8 0

2 years ago

Considerando que você comece a caminhar em velocidade constante, inicialmente a 350 m de um ponto referencial escolhido. Você ca

ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

$S(t)=S_o+vt$ ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

$S(t)=350-(1m/s)t$

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

(1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

4 0

2 years ago