6) Magnetic flux density is per unit area

Simora [160]

I only know what number 1. is and its Mechanical Energy.

8 0

3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

5 0

4 years ago

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration th

vekshin1

the average acceleration of the bullet through the board -55657×10⁵ m/s²

acceleration- Rate of change of velocity with respect to time.

S = 0.10 mm = 10⁻⁵ m ( distance) [∵ 1mm = 10⁻³]

u = 480 m/s (initial velocity)

v = 345 m/s (final velocity)

As we know the 3rd equation of motion.

v² = u² + 2aS

a = ? ( acceleration)

using these values in equation we get

(345)² = (480)² + 2×a× 10⁻⁵

a = (345)² - (480)² / 2× 10⁻⁵

a = -55657×10⁵ m/s²

The negative sign shows that the direction of acceleration is opposite of velocity thus bullet is slowing down.

the average acceleration of the bullet through the board -55657×10⁵ m/s²

The given question is incomplete. The complete question is.

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration through the board, and emerges with a speed of 345 m/sm/s.

What is the average acceleration of the bullet through the board/?

to know more about the equation of motion :

brainly.com/question/13269040

#SPJ4

5 0

2 years ago

A spring-launched toy is propelled into the air. At first when the spring is compressed and the toy is placed on top. Before the

wlad13 [49]

In this problem, you are asked to find a vertical position of a ball when you are given its initial position on a spring. In both locations, the speed of the ball is zero.

If non-conservative forces are either known or small and if energy is converted from one form to another between the locations, then any time you relate speed and position of an object at two different points, conservation of energy is the most direct way to understand the problem.

In this case, you start out with stored energy in the compression of the spring and convert it to stored gravitational energy.

3 0

3 years ago

Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to th

Nastasia [14]

Answer:

$T = 40 + e^{3.68t} e^{2.99}$